Which of the following units is useful in relating concentration of solution with its vapour pressure? (i) mol

NCERT Exemplar · Q1

Q. Which of the following units is useful in relating concentration of solution with its vapour pressure?

(i) mole fraction

(ii) parts per million

(iii) mass percentage

(iv) molality

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Concept understanding — Mass Percentage

Mass Percentage: The Intuition

Imagine you're making lemonade. You mix 50 grams of sugar into 200 grams of water. The total drink weighs 250 grams. Now, if someone asks, "How much of this drink is actually sugar?" — you're not just saying "50 grams." You want to say what fraction of the whole mixture is sugar, scaled to a convenient 100.

That's mass percentage. It answers: "Out of every 100 grams of the mixture, how many grams are this particular component?"

In our lemonade, sugar is 50 g out of 250 g total. That's $\frac{50}{250} = 0.2$ of the whole. Multiply by 100 to get the percentage: $0.2 \times 100 = 20\%$ . So, 20% of the drink's mass is sugar. If you had 100 g of this lemonade, 20 g of it would be sugar.

The Precise Definition

\text{Mass percentage of component} = \frac{\text{Mass of that component}}{\text{Total mass of mixture}} \times 100\%

The formula is simple, but the key is understanding what "total mass" means. It's the sum of masses of all components in the mixture — nothing more, nothing less.

Why It Matters in Chemistry

Mass percentage is one of the most common ways to express concentration — how much of a substance is present in a mixture. You'll see it in:

Solutions: "10% salt water" means 10 g of salt dissolved in enough water to make 100 g of solution (not 10 g salt + 100 g water — that would be 110 g total, giving only about 9.1%).
Alloys: "18-karat gold" is 75% gold by mass (18 parts gold out of 24 total parts).
Food labels: "Fat: 15%" means 15 g of fat per 100 g of the food.

Watch out

A Common Mistake

Students often think "10% salt solution" means 10 g salt + 100 g water. That's wrong. It means 10 g salt + 90 g water = 100 g total solution. The denominator is total mass, not the mass of the solvent alone.

Step-by-Step Example

Problem: A solution is made by dissolving 25 g of glucose in 175 g of water. Find the mass percentage of glucose.

Step 1: Identify the component you care about — glucose (25 g).

Step 2: Find the total mass of the mixture.

\text{Total mass} = 25\ \text{g (glucose)} + 175\ \text{g (water)} = 200\ \text{g}

Step 3: Apply the formula.

\text{Mass percentage of glucose} = \frac{25}{200} \times 100\% = 12.5\%

Interpretation: In every 100 g of this solution, 12.5 g is glucose and the rest (87.5 g) is water.

When to Use Mass Percentage vs. Other Measures

Mass percentage is ideal when:

You're working with solid mixtures or solutions where masses are easy to measure.
You want a concentration that doesn't change with temperature (unlike volume-based measures like molarity, which expand/contract with heat).

It's less useful when you need to count molecules (use mole fraction) or when volumes are more practical (use volume percentage).

One Final Check

If you ever get confused, go back to the lemonade. The question is always: "What fraction of the total weight is this one thing?" Multiply that fraction by 100, and you have your mass percentage.

Queries such as "mass percentage formula chemistry" and "mass percentage class 12 solutions" are common around this topic, which is a core concentration term introduced in the Solutions chapter of the NCERT/CBSE Class 12 Chemistry curriculum. Distinguishing it correctly from mass/volume percentage is a frequent numerical-question type in board exams and JEE Main.

Why this formula?

Let's break down Mass Percentage from first principles. The goal is to understand why the formula is what it is, not just to memorize it.

1. The Core Idea: "Part of a Whole"

Mass percentage answers a simple question: "If I break a mixture into 100 equal parts by mass, how many of those parts come from a specific component?"

Imagine you have a bowl of fruit salad. The total mass is 500 grams. The apples in it weigh 100 grams.

The apples are a part of the whole salad.
The whole salad is the total.

The mass percentage tells you the fraction of the total mass that is apples, but expressed "out of 100" (per cent).

2. The Natural First Step: The Fraction

Before we talk about "percentage," we talk about the fraction of the total:

\text{Fraction of component} = \frac{\text{Mass of component}}{\text{Total mass of mixture}}

For the apple example:

\frac{100 \text{ g}}{500 \text{ g}} = 0.2

This means 0.2 (or one-fifth) of the total mass is apples. This is the pure ratio — no scaling yet.

3. Why Multiply by 100?

A fraction like 0.2 is perfectly correct, but it's not intuitive for quick comparison. "Per cent" literally means "per hundred" (from Latin per centum).

To convert a fraction into a "per hundred" number, we multiply by 100:

\text{Percentage} = (\text{Fraction}) \times 100

So:

0.2 \times 100 = 20\%

This tells us: "Out of every 100 grams of fruit salad, 20 grams come from apples." That's much easier to visualize.

4. The Final Formula (The "Why" in One Line)

Putting the fraction and the "times 100" together gives the standard formula:

\text{Mass percentage} = \frac{\text{Mass of component}}{\text{Total mass of mixture}} \times 100\%

Why does this work?

Because it's just:

Find the proportion (part ÷ whole).
Scale that proportion to per hundred (× 100).

5. A Common Exam Trap (and Why It's Wrong)

Sometimes students write:

\text{Mass percentage} = \frac{\text{Mass of component}}{\text{Mass of solvent}} \times 100

This is incorrect. Why?

The denominator must be the total mass of the entire mixture (solute + solvent), not just the solvent.
The percentage tells you the share of the whole, not the share of one part relative to another.

Correct example:

10 g salt in 90 g water → total = 100 g.

Mass % of salt = $\frac{10}{100} \times 100 = 10\%$ (not $\frac{10}{90} \times 100 \approx 11.1\%$ ).

6. Quick Summary for Exams

Step	What to do	Why
1	Find the mass of the component	It's the "part"
2	Find the total mass of the mixture	It's the "whole"
3	Divide part by whole	Gives the fraction
4	Multiply by 100	Converts fraction to "per hundred"

Final takeaway: Mass percentage is just a scaled fraction — it makes comparisons easy by always using a base of 100.

The key idea is that Raoult’s law directly relates vapour pressure to the mole fraction of the solvent (or solute). Among the given options, only mole fraction appears in the law itself - the others are indirect or irrelevant. The correct answer is (i) mole fraction.

Why this question matters

When you study solutions and their colligative properties, the link between concentration and vapour pressure is fundamental. Raoult’s law states that the vapour pressure of a solvent above a solution equals the product of its mole fraction in the solution and its vapour pressure in the pure state:

P_{\text{solvent}} = x_{\text{solvent}} \cdot P_{\text{solvent}}^\circ

This is a direct, proportional relationship. The question asks which concentration unit is useful in relating concentration to vapour pressure - meaning which one appears naturally in the law itself.

Step-by-step reasoning

Recall Raoult’s law. For a solution of a non-volatile solute in a volatile solvent, $\frac{P^\circ - P}{P^\circ} = x_{\text{solute}}$ . No other concentration unit appears in this equation.
Examine each option
- (i) Mole fraction - dimensionless, appears directly in Raoult’s law. The natural variable for vapour-pressure relations.
- (ii) Parts per million (ppm) - a mass/volume-based ratio; does not appear in any vapour-pressure equation.
- (iii) Mass percentage - can be converted to mole fraction, but is not itself used in Raoult’s law.
- (iv) Molality - useful for boiling-point elevation and freezing-point depression, but not for vapour pressure.
Why the others are not “useful” in this context. Only mole fraction appears directly in the mathematical relationship; the others require conversion first.

Watch out

A common mistake is to pick molality because it is used for other colligative properties. But vapour-pressure lowering is directly proportional to mole fraction, not molality.

✓Final answer

The correct option is (i) mole fraction.

Method: Raoult's Law & Vapour Pressure Relation

The correct answer is (i) mole fraction.

Why Mole Fraction?

Vapour pressure of a solution is directly related to the mole fraction of the solvent via Raoult's Law:

P_{\text{solution}} = x_{\text{solvent}} \cdot P_{\text{solvent}}^\circ

where:

$P_{\text{solution}}$ = vapour pressure of the solution
$x_{\text{solvent}}$ = mole fraction of the solvent
$P_{\text{solvent}}^\circ$ = vapour pressure of pure solvent

Why Not the Others?

Unit	Reason it doesn't directly relate to vapour pressure
Parts per million (ppm)	Mass-based ratio; no direct link to mole fraction in Raoult's Law
Mass percentage	Also mass-based; doesn't appear in vapour pressure equations
Molality	Temperature-independent but still mass-based; not directly in Raoult's Law

Key Takeaway

Mole fraction is the only concentration unit that appears directly in Raoult's Law, making it the natural choice for relating concentration to vapour pressure.

\boxed{\text{(i) mole fraction}}

Correct Answer

(i) mole fraction

Why? Raoult's law states that the vapour pressure of a solution is directly proportional to the mole fraction of the solvent. Mathematically:

P_{\text{solution}} = X_{\text{solvent}} \cdot P_{\text{solvent}}^\circ

No other concentration unit appears directly in this law.

Common Mistakes & How to Avoid Them

1. Choosing mass percentage (option C)

The mistake: Students think "mass percentage" is the most common concentration unit, so it must relate to vapour pressure.

Why it's wrong: Mass percentage tells you grams of solute per 100 g of solution. Vapour pressure depends on the number of particles (moles) in the solution, not their mass. Two solutions with the same mass percentage can have very different vapour pressures if the solutes have different molar masses.

How to avoid: Always ask: "Does this unit count particles or just mass?" For vapour pressure, you need a particle-counting unit.

2. Choosing molality (option D)

The mistake: Students recall that molality is used in colligative properties (like boiling point elevation) and assume it works for vapour pressure too.

Why it's wrong: Molality ( $m$ ) is moles of solute per kg of solvent. While it is a particle-counting unit, Raoult's law uses mole fraction, not molality. Molality is useful for boiling point and freezing point, but not directly for vapour pressure.

How to avoid: Memorise the specific formula for each colligative property:

Vapour pressure → mole fraction
Boiling point elevation / freezing point depression → molality
Osmotic pressure → molarity

3. Choosing parts per million (option B)

The mistake: Students think "ppm is very precise, so it must be useful for vapour pressure."

Why it's wrong: ppm is just a scaled-up version of mass percentage (mg per kg). It still ignores particle count. It's used for trace concentrations (pollutants, minerals), not for vapour pressure calculations.

How to avoid: Remember that ppm, mass percentage, and volume percentage are all mass-based or volume-based units. Vapour pressure is a particle-based property.

4. Confusing mole fraction with mass fraction

The mistake: Students know "fraction" is involved, but they calculate mass fraction instead of mole fraction.

Example: For a solution of 10 g NaCl in 90 g water:

Mass fraction of NaCl = $10/100 = 0.1$
Mole fraction of NaCl = $\frac{10/58.5}{(10/58.5) + (90/18)} \approx 0.033$

These are very different — using mass fraction in Raoult's law gives a wrong answer.

How to avoid: Always convert given masses to moles before calculating mole fraction. Never substitute mass fraction directly.

Quick Summary Table

Unit	Counts particles?	Used in Raoult's law?
Mole fraction	✓ Yes	✓ Yes
Molality	✓ Yes	✗ No (used for $\Delta T_b$ , $\Delta T_f$ )
Mass percentage	✗ No	✗ No
ppm	✗ No	✗ No

Final tip: When you see "vapour pressure" in a question, immediately think mole fraction — it's the only unit that appears in Raoult's law directly.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 28 on this concept.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$N = 6 \times 10^{23}\ mol^{-1}$
›Reveal solutionSolution
A straightforward percentage-composition → moles → atoms calculation; the answer is $1.2\times10^{21}$ Mg atoms.

Concept and Intuition

Percentage composition tells us the mass fraction of an element in a compound. Once we know the mass of that element, dividing by its atomic mass gives moles, and multiplying by Avogadro's number gives the actual atom count — the standard mass→mole→number chain.

Step-by-Step Solution

Mass of Mg in 2.0 g chlorophyll $= 2.4\% \times 2.0\ g = 0.024 \times 2.0 = 0.048$ g.

Moles of Mg $= \dfrac{0.048}{24} = 0.002$ mol.

Number of Mg atoms $= 0.002 \times 6\times10^{23} = 1.2\times10^{21}$ .

Common Mistakes

Using the percentage directly as a mole fraction instead of converting to mass first.

Arithmetic slips in the division by atomic weight.

✓Final answer
$1.2\times10^{21}$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$g\,mol^{-1}$
›Reveal solutionSolution
Mustard gas, $(ClCH_2CH_2)_2S$ ( $C_4H_8Cl_2S$ , $M=159$ ), is 20.1% S and 44.65% Cl by mass.

Concept and Intuition

Percentage composition just needs the correct molecular formula and the atomic masses of the elements asked about, divided by the total molar mass.

Step-by-Step Solution

Mustard gas is bis(2-chloroethyl) sulfide, $(ClCH_2CH_2)_2S$ , i.e. $C_4H_8Cl_2S$ .

Molar mass check: $C_4=48$ , $H_8=8$ , $Cl_2=2\times35.5=71$ , $S=32$ . Sum $=48+8+71+32=159\,g\,mol^{-1}$ — matches the given value, confirming the formula.

Mass fraction of S $=\dfrac{32}{159}=0.2013 \Rightarrow 20.12\%$ (rounded as in the option).

Mass fraction of Cl $=\dfrac{71}{159}=0.4465 \Rightarrow 44.65\%$ .

So %S, %Cl (respectively) $=20.12,\ 44.65$ .

Common Mistakes

Using the wrong number of Cl atoms (there are two, not one) and getting a Cl percentage close to 22% instead of 44.65%.

Swapping S and Cl percentages — the question explicitly asks "S then Cl", matching option (A), not (B).

✓Final answer
$The correct option is (A) — 20.12, 44.65 (percentage of S, then Cl).$

ANSWER: A
AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$27°C$
›Reveal solutionSolution
Compute the mixture's average molar mass from the mass percentages, then apply the ideal-gas density formula $d=PM/RT$ to get $\approx 0.77\ \text{g L}^{-1}$ .

Concept and Intuition

For a gas mixture, the effective (average) molar mass is the total mass divided by total moles of all species combined. Once you have that average $M$ , the density of the mixture behaves just like a single ideal gas of that molar mass: $d = PM/RT$ .

Step-by-Step Solution

Assume 100 g of mixture: mass of $\mathrm{CH_4} = 64$ g, mass of $\mathrm{C_2H_6} = 36$ g.

Moles of $\mathrm{CH_4}$ ( $M=16$ g/mol) $= 64/16 = 4$ mol.

Moles of $\mathrm{C_2H_6}$ ( $M=30$ g/mol) $= 36/30 = 1.2$ mol.

Total moles $= 4+1.2 = 5.2$ mol; total mass $=100$ g.

Average molar mass $M_{avg} = 100/5.2 = 19.23$ g/mol.

Convert pressure: $P = 750/760 = 0.9868$ atm; $T = 27°C = 300$ K.

$d = \dfrac{PM}{RT} = \dfrac{0.9868 \times 19.23}{0.082 \times 300} = \dfrac{18.98}{24.6} \approx 0.77\ \text{g/L}$ .

Common Mistakes

Using $R=0.0821$ inconsistently or forgetting to convert mmHg to atm.

Computing a mole-weighted average incorrectly (e.g., averaging mass % directly instead of via moles).

✓Final answer
$The correct option is (D) — 0.77.$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\mathrm{H_2}$
›Reveal solutionSolution
Tests converting partial-pressure (mole fraction) data into a weight percentage using molar masses; the answer is 33.33%.

Concept and Intuition

Partial pressure is proportional to mole fraction (Dalton's law: $p_i = x_i P_{total}$ at constant T, V). Once we know the mole fractions of each gas, we can find the mass of each component (for an assumed total of 1 mole of gas mixture) using their molar masses, and hence the weight percentage.

Step-by-Step Solution

Mole fraction of $H_2$ : $x_{H_2} = \dfrac{p_{H_2}}{P_{total}} = \dfrac{1.778}{2} = 0.889$ .

Mole fraction of $O_2$ : $x_{O_2} = 1 - 0.889 = 0.111$ .

Assume 1 total mole of gas mixture. Mass of $H_2 = 0.889 \times 2\ \text{g/mol} = 1.778$ g. Mass of $O_2 = 0.111 \times 32\ \text{g/mol} = 3.552$ g.

Total mass = $1.778 + 3.552 = 5.33$ g.

Weight % of $H_2 = \dfrac{1.778}{5.33}\times 100 \approx 33.33\%$ .

Common Mistakes

Confusing mole percentage with weight percentage — since $O_2$ is 16 times heavier per mole than $H_2$ , even a small mole fraction of $O_2$ contributes a large mass fraction.

Forgetting to normalize by total mass rather than just reporting the mole fraction as the answer.

✓Final answer
$The correct option is (B) — 33.33.$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\mathrm{CaCO_3}$
›Reveal solutionSolution
Tests mass-balance on a two-component thermal decomposition mixture; solving simultaneous equations gives 54.34% $CaCO_3$ .

Concept and Intuition

Both carbonates decompose on strong heating: $CaCO_3 \to CaO + CO_2$ and $MgCO_3 \to MgO + CO_2$ , each losing $CO_2$ mass. Since the two salts lose mass in different fixed ratios (dictated by their molar masses), the residue mass depends on the mixture's composition — letting us solve for the unknown split.

Step-by-Step Solution

Let mass of $CaCO_3 = x$ g, so mass of $MgCO_3 = (1.84-x)$ g.

$CaCO_3$ (M = 100 g/mol) → $CaO$ (M = 56 g/mol): mass of CaO produced $= \dfrac{56}{100}x = 0.56x$ .

$MgCO_3$ (M = 84 g/mol) → $MgO$ (M = 40 g/mol): mass of MgO produced $= \dfrac{40}{84}(1.84-x)$ .

Total residue: $0.56x + \dfrac{40}{84}(1.84-x) = 0.96$ .

Compute $\dfrac{40}{84} = 0.4762$ : $0.56x + 0.4762(1.84) - 0.4762x = 0.96$ $\Rightarrow 0.56x + 0.8762 - 0.4762x = 0.96$ $\Rightarrow 0.0838x = 0.0838$ $\Rightarrow x = 1.0$ g.

Percentage of $CaCO_3 = \dfrac{1.0}{1.84}\times 100 = 54.35\% \approx 54.34\%$ .

Common Mistakes

Using the wrong decomposition ratio (e.g., mixing up CaO/CaCO3 with MgO/MgCO3 ratios).

Forgetting both salts lose mass — assuming only one component decomposes.

✓Final answer
$The correct option is (C) — 54.34.$

ANSWER: C
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$Q.The mole fractions of glucose and water in aqueous glucose solution are 0.0244 and 0.9756 respectively. What is the weight percentage (w/w) of glucose in this solution ? (A) 40 (B) 25 (C) 20 (D) 10$
›Reveal solutionSolution
Converting mole fraction to weight percentage for a glucose–water solution gives approximately 20% w/w glucose.

Concept and Intuition

Mole fraction tells us the mole ratio of components; to get weight percentage we must convert moles to mass using the molar masses (glucose $M = 180\ \mathrm{g/mol}$ , water $M = 18\ \mathrm{g/mol}$ ), then take the mass fraction.

Step-by-Step Solution

Assume a total of 1 mole of solution (mole fractions given directly as fractions of 1 mole total).

Moles of glucose $= 0.0244$ , moles of water $= 0.9756$ .

Mass of glucose $= 0.0244 \times 180 = 4.392\ \mathrm{g}$ .

Mass of water $= 0.9756 \times 18 = 17.5608\ \mathrm{g}$ .

Total mass of solution $= 4.392 + 17.5608 = 21.9528\ \mathrm{g}$ .

Weight percentage of glucose $= \dfrac{4.392}{21.9528}\times 100 \approx 20.01\% \approx 20\%$ .

Common Mistakes

Using the mole fraction directly as a weight fraction without converting through molar mass.

Forgetting glucose's molar mass is 180 g/mol (C $_6$ H $_{12}$ O $_6$ ), not 342 (that's sucrose).

✓Final answer
$The correct option is (C) — 20.$

ANSWER: C
AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$\mathrm{H_2}$
›Reveal solutionSolution
Converting the given weight percentage to moles (using molar masses 2 and 32 g/mol) gives a partial pressure ratio of $8:1$ .

Concept and Intuition

By Dalton's law of partial pressures, for a gas mixture at a common temperature and volume, the partial pressure of each component is proportional to its mole fraction (equivalently, to its number of moles). So converting mass percentages to moles (using each gas's molar mass) directly gives the pressure ratio.

Step-by-Step Solution

Take a convenient total mass, say $3\ \text{g}$ , so that $33.33\%=1/3$ works out to whole numbers: mass of H $_2$ $=1\ \text{g}$ , mass of O $_2$ $=2\ \text{g}$ .

Moles of H $_2$ : $n_{H_2} = \frac{1}{2} = 0.5\ \text{mol}$

Moles of O $_2$ : $n_{O_2} = \frac{2}{32} = 0.0625\ \text{mol}$

Ratio of moles (= ratio of partial pressures, by Dalton's law at fixed $T,V$ ): $\frac{n_{H_2}}{n_{O_2}} = \frac{0.5}{0.0625} = 8$

So $P_{H_2}:P_{O_2} = 8:1$ .

Common Mistakes

Using the mass ratio (1:2) directly as the pressure ratio, forgetting to convert to moles via molar mass.

Using the wrong molar mass for O $_2$ (must be 32 g/mol, not 16).

✓Final answer
$The correct option is (A) — 8 : 1.$

ANSWER: A
AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$Q.What are the mole fractions of glucose and water respectively, in 20% (w/w) aqueous glucose solution ? (C = 12 u; O = 16 u; H = 1 u) (A) 0.0244, 0.9756 (B) 0.04, 0.96 (C) 0.0636, 0.9364 (D) 0.0124, 0.9876$
›Reveal solutionSolution
Taking a 100 g basis of the 20% w/w solution gives 20 g glucose and 80 g water; converting to moles and dividing gives mole fractions 0.0244 (glucose) and 0.9756 (water).

Concept and Intuition

"20% w/w" means 20 g of solute per 100 g of solution, so the remaining 80 g is solvent. Mole fraction of each component is its moles divided by the total moles of all components.

Step-by-Step Solution

Basis: 100 g solution → 20 g glucose ( $\mathrm{C_6H_{12}O_6}$ , $M=180$ g/mol), 80 g water ( $M=18$ g/mol).

Moles of glucose $= 20/180 = 0.1111$ mol.

Moles of water $= 80/18 = 4.4444$ mol.

Total moles $= 0.1111 + 4.4444 = 4.5556$ mol.

Mole fraction of glucose $= 0.1111/4.5556 = 0.0244$ .

Mole fraction of water $= 4.4444/4.5556 = 0.9756$ .

Check: $0.0244+0.9756=1.000$ ✓.

Common Mistakes

Using the wrong molar mass for glucose (it is $\mathrm{C_6H_{12}O_6}=180$ g/mol, not 90 or 342 as in sucrose).

Forgetting mole fraction uses moles, not mass fractions directly.

✓Final answer
$The correct option is (A) — 0.0244, 0.9756.$

ANSWER: A
AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$Q.In aqueous glucose solution, the mole fraction of water is 40 times to mole fraction of glucose. What is the weight percentage (w/w) of glucose in the solution? (A) 40 (B) 30 (C) 20 (D) 10$
›Reveal solutionSolution
This tests converting mole-fraction ratio into moles, then into a weight percentage. The answer is 20%.

Concept and Intuition

Mole fraction compares moles of a component to total moles, while weight percentage compares mass of a component to total mass. To go from one to the other we must first fix actual mole amounts (using the given ratio), then convert moles to grams using molar masses (glucose $M=180\ \mathrm{g\,mol^{-1}}$ , water $M=18\ \mathrm{g\,mol^{-1}}$ ).

Step-by-Step Solution

Let mole fraction of glucose $=x_2$ and of water $=x_1$ . Given $x_1 = 40x_2$ .

Since $x_1+x_2=1$ : $40x_2+x_2=1 \Rightarrow x_2=\dfrac{1}{41},\ x_1=\dfrac{40}{41}$ .

Take $n_2=1$ mol glucose $\Rightarrow n_1=40$ mol water (same ratio as mole fractions since total moles cancel).

Mass of glucose $=1\times180=180$ g. Mass of water $=40\times18=720$ g.

Total mass of solution $=180+720=900$ g.

Weight percentage of glucose $=\dfrac{180}{900}\times100=20\%$ .

Common Mistakes

Confusing mole fraction directly with weight fraction — they are equal only when molar masses are equal, which is not the case here.

Forgetting to multiply moles by the correct molar mass (180 for glucose, 18 for water).

✓Final answer
$The correct option is (C) — 20.$

ANSWER: C
AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\mathrm{NaCl}$
›Reveal solutionSolution
This is a mixture stoichiometry problem: using the given %Na by mass to set up and solve a linear equation for the NaCl fraction, giving 45%.

Concept and Intuition

Each mole of NaCl and each mole of NaBr contributes exactly one mole of Na. So if we know the total mass of Na present, and express the unknown masses of NaCl and NaBr in terms of one variable, we can solve for the composition directly using molar masses.

Step-by-Step Solution

Total mixture mass = 4.0 g. Na is 30% of this by mass: mass of Na $= 0.30 \times 4.0 = 1.2$ g.

Moles of Na $= \dfrac{1.2}{23} = 0.052174$ mol.

Let mass of NaCl $= x$ g, so mass of NaBr $= (4.0 - x)$ g.

Molar mass of NaCl $= 23+35.5 = 58.5$ ; molar mass of NaBr $= 23+80=103$ .

Moles of Na from both salts: $\dfrac{x}{58.5} + \dfrac{4.0-x}{103} = 0.052174$ .

Multiply through by $58.5\times103 = 6025.5$ : $103x + 58.5(4.0-x) = 314.35$ .

$103x + 234 - 58.5x = 314.35 \Rightarrow 44.5x = 80.35 \Rightarrow x = 1.806$ g.

%NaCl by mass $= \dfrac{1.806}{4.0}\times100 \approx 45.1\%$ .

Common Mistakes

Forgetting Na comes from both salts and only crediting it to NaCl.

Arithmetic slip in setting up the mass balance equation with the two different molar masses.

✓Final answer
$The correct option is (C) — 45%.$

ANSWER: C
AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$\mathrm{NH_2CONH_2}$
›Reveal solutionSolution
Using the solution's density to get the total mass per litre, subtracting the urea mass gives the water mass; converting both to moles gives a urea mole fraction of about 0.038.

Concept and Intuition

Mole fraction requires the moles of each component in a defined amount of solution. Density lets us convert the given volume (1 dm³ = 1 L) into total solution mass, from which subtracting the known solute mass gives the solvent (water) mass.

Step-by-Step Solution

Total solution mass in 1 dm³ = density × volume = $1.02\ \mathrm{kg/dm^3}\times1\ \mathrm{dm^3} = 1020$ g.

Mass of urea = 120 g (given), so mass of water = $1020-120 = 900$ g.

Molar mass of urea, $\mathrm{NH_2CONH_2}$ : $2(14)+4(1)+12+16 = 28+4+12+16=60$ g/mol.

Moles of urea $= 120/60 = 2$ mol.

Moles of water $= 900/18 = 50$ mol.

Mole fraction of urea $= \dfrac{2}{2+50} = \dfrac{2}{52} = 0.0385 \approx 0.038$ .

Common Mistakes

Using the volume (1 L) directly as the water mass (1000 g), ignoring the actual solution density and the fact that water mass = total solution mass − solute mass.

Miscalculating the molar mass of urea (a frequent slip is missing one of the two N or forgetting the carbonyl O).

✓Final answer
$The correct option is (A) — 0.038.$

ANSWER: A
AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
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›Reveal solutionSolution
Converting 1000 ppm of CaCO $_3$ (≈ 1 g/L for a dilute aqueous solution) into molarity using M = 100 g/mol gives $10^{-2}$ mol/L — option (D).

Concept and Intuition

"ppm" (parts per million) for a dilute aqueous solution is conventionally taken as mg of solute per litre of solution (since the solution's density is essentially that of water, 1 g/mL, at these very low concentrations). Once we have the mass concentration in g/L, dividing by the molar mass converts it to molarity directly.

Step-by-Step Solution

1000 ppm = 1000 mg solute per L of solution = 1 g/L (for a dilute aqueous solution where density ≈ 1 g/mL).

Molar mass of $\mathrm{CaCO_3}$ : $40 (\text{Ca}) + 12 (\text{C}) + 3\times16 (\text{O}) = 40+12+48 = 100$ g/mol.

Molarity $= \dfrac{\text{mass}/\text{L}}{\text{molar mass}} = \dfrac{1\ \text{g/L}}{100\ \text{g/mol}} = 0.01\ \text{mol/L} = 10^{-2}\ \text{mol/L}$ .

Common Mistakes

Misremembering ppm as parts per thousand or using a wrong conversion factor (1000 ppm = 1000 mg/L = 1 g/L, not 1 mg/L).

Using the wrong molar mass for CaCO $_3$ (forgetting the three oxygens).

✓Final answer
$10^{-2}$

ANSWER: D

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Q.Which of the following units is useful in relating concentration of solution with its vapour pressure?

Why this question matters

Step-by-step reasoning

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