What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water? (i) 0.1 m (ii) 1 M

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Q. What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water?

(i) 0.1 m

(ii) 1 M

(iii) 0.5 m

(iv) 1 m

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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.

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

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: …

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Molality is defined as moles of solute per kilogram of solvent. By calculating the moles of HCl and converting the mass of water to kilograms, we find the molality to be $\boxed{\text{1 m}}$ .

Molality is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent. It's a particularly useful concentration unit because, unlike molarity, it does not depend on temperature. Volume changes with temperature, but mass does not, making molality a temperature-independent measure of concentration.

m

Let's apply this concept to the given problem.

Identify the given quantities:

We are given the mass of the solute, HCl gas, and the mass of the solvent, water.
- Mass of HCl (solute) $= 18.25 \text{ g}$
- Mass of water (solvent) $= 500 \text{ g}$
Calculate the moles of solute (HCl):

To find the moles of HCl, we need its molar mass.
- Molar mass of H $= 1.008 \text{ g/mol}$
- Molar mass of Cl $= 35.45 \text{ g/mol}$
- Molar mass of HCl $= 1.008 + 35.45 = 36.458 \text{ g/mol}$ For practical calculations in such problems, it's common to use rounded values like $36.5 \text{ g/mol}$ for HCl, which simplifies the calculation significantly and is often implied by the numbers given. Let's proceed with $36.5 \text{ g/mol}$ .
Moles of HCl $= \frac{\text{Mass of HCl}}{\text{Molar mass of HCl}}$

Moles of HCl $= \frac{18.25 \text{ g}}{36.5 \text{ g/mol}} = 0.5 \text{ mol}$
Convert the mass of solvent to kilograms:

The definition of molality requires the mass of the solvent to be in kilograms.

Mass of water $= 500 \text{ g}$

Mass of water in kg $= 500 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 0.5 \text{ kg}$
Calculate the molality of the solution:

Now we have the moles of solute and the mass of solvent in kilograms, so we can directly apply the molality formula. …

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Here are the common mistakes students make when solving this problem, along with the correct reasoning to avoid them.

Mistake 1: Confusing Molality with Molarity

The Error: Students see "HCl gas in water" and immediately calculate molarity (M). They take the volume of water (500 mL) as the volume of the solution.
- They calculate moles of HCl: $18.25 / 36.5 = 0.5$ moles.
- They then do: $0.5 / 0.5 = 1 \text{ M}$ .
- They pick option (ii) 1 M or mistakenly think $1 \text{ M} = 1 \text{ m}$ .
Why it's wrong: Molality ( $m$ ) uses mass of solvent in kg, not volume of solution. Molarity uses volume of solution. Water's density is ~1 g/mL, but the final solution volume is not exactly 500 mL. The question explicitly asks for molality.
How to avoid: Read the unit symbol carefully. "m" means molal (mol/kg). "M" means molar (mol/L). Always check if the question asks for molality or molarity before starting.

Mistake 2: Using the Wrong Mass for the Solvent

The Error: Students include the mass of the solute (HCl) in the mass of the solvent.
- They calculate: Mass of solution = $18.25 + 500 = 518.25 \text{ g}$ .
- They then use $0.51825 \text{ kg}$ as the solvent mass.
Why it's wrong: Molality is defined as moles of solute per kilogram of solvent (the substance doing the dissolving). The solute (HCl) is dissolved in the water. The water is the solvent.
How to avoid: Identify the solvent first. In an aqueous solution, water is almost always the solvent. The mass of the solvent is given directly: 500 g of water. Convert that to kg: $500 \text{ g} = 0.5 \text{ kg}$ .

Mistake 3: Incorrect Molar Mass Calculation

The Error: Students use the wrong molar mass for HCl.
- They might use $H = 1$ , $Cl = 35$ (instead of 35.5), giving $36 \text{ g/mol}$ .
- Or they might use $Cl = 35.5$ but forget to add H, using $35.5 \text{ g/mol}$ .
Why it's wrong: The atomic mass of chlorine is 35.5 u (average of isotopes). Using 35 or 36 changes the mole calculation significantly.
How to avoid: Memorize common atomic masses. For HCl: $H = 1$ , $Cl = 35.5$ , so molar mass = $1 + 35.5 = 36.5 \text{ g/mol}$ . Always write it down before calculating moles. …

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

COMEDK 2026Set 2026-M1 markMCQ
$X$
›Reveal solutionSolution
Mass percent is the mass of solute divided by the total mass of solution, times 100. Here, the solute mass is 4.0 g, and the solvent (water) mass is 2.0 moles × 18 g/mol = 36 g, so total mass = 40 g, giving mass percent = (4/40)×100 = 10%. The correct option is (D).

Concept & Intuition

Mass percent tells you how many grams of solute are present in every 100 grams of solution. It’s a simple ratio:
$\text{mass percent} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$
The trick here is that the solvent (water) is given in moles, not grams. So the first step is always to convert moles of water to grams using its molar mass (18 g/mol). Once everything is in grams, the calculation is straightforward.

Step-by-step solution

Find the mass of water (solvent) We have 2.0 moles of water. The molar mass of water is $18.0\ \text{g/mol}$ .

$\text{mass of water} = 2.0\ \text{mol} \times 18.0\ \frac{\text{g}}{\text{mol}} = 36\ \text{g}$

Find the total mass of the solution The solution contains the solute (4.0 g of X) plus the solvent (36 g of water).

$\text{total mass} = 4.0\ \text{g} + 36\ \text{g} = 40\ \text{g}$

Calculate the mass percent
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KCET 2025Set D-41 markMCQ
$Q.Which of the following methods of expressing concentration are unitless? (A) Mole fraction and Mass percent (W/W) (B) Molality and Mole fraction (C) Mass percent (W/W) and Molality (D) Molality and Molarity$
›Reveal solutionSolution
A concentration term is unitless only when it is a ratio of two quantities of the same kind — mole/mole or mass/mass — so the units cancel.

Step 1 — Write each concentration measure with its units.

Mole fraction

$x_A = \frac{n_A}{n_A + n_B} = \frac{\mathrm{mol}}{\mathrm{mol}}$
Moles divided by moles $\Rightarrow$ unitless (and it always lies between 0 and 1).

Mass percent (W/W)

$\%\,(w/w) = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 = \frac{\mathrm{g}}{\mathrm{g}} \times 100$
Grams divided by grams $\Rightarrow$ unitless (the "%" is a pure number, not a unit).

Molality

$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \mathrm{mol\,kg^{-1}}$
Moles divided by mass — two different kinds of quantity $\Rightarrow$ has units.

Molarity

$M = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \mathrm{mol\,L^{-1}}$
Moles divided by volume $\Rightarrow$ has units.

Step 2 — Apply the test.

The unitless pair is therefore mole fraction and mass percent (W/W).

Step 3 — Eliminate.

(B) Molality has units ( $\mathrm{mol\,kg^{-1}}$ ) — fails.

(C) Molality has units — fails. …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{Cl}=35.5 \mathrm{~g} / \mathrm{mol}$
›Reveal solutionSolution
The key is to use the given chlorine mass percentage to find the molar mass of the haloalkane, then compute the number of molecules in 0.1 g, and finally multiply by one Cl atom per molecule. The result is about $9.328 \times 10^{20}$ Cl atoms, so the correct option is (C).

Concept & Intuition

We have a haloalkane (an alkane with one chlorine atom replacing a hydrogen). The problem tells us that chlorine makes up 55% of the mass of one molecule. That means if we know the mass of one mole of the compound, we can find how many moles of Cl are in a sample. Since each molecule has exactly one Cl atom, the number of Cl atoms equals the number of molecules. So the plan: find the molar mass from the percentage, then convert 0.1 g to moles, then to atoms via Avogadro’s number.

Step-by-step solution

Relate percentage to molar mass Let $M$ be the molar mass of the haloalkane (in g/mol). One mole of the compound contains one mole of chlorine atoms, which has mass $35.5$ g. The percentage by mass of chlorine is:

$\frac{35.5}{M} \times 100\% = 55\%$
So:
$\frac{35.5}{M} = 0.55$
Solving:
$M = \frac{35.5}{0.55} = 64.545\ldots \ \text{g/mol}$
(We can keep it as $\frac{35.5}{0.55}$ for now.)

Find moles of haloalkane in 0.1 g Moles of compound:

$n = \frac{\text{mass}}{M} = \frac{0.1}{35.5/0.55} = \frac{0.1 \times 0.55}{35.5}$
Simplify:
$n = \frac{0.055}{35.5} \ \text{mol}$

Number of molecules (and thus Cl atoms) Since each molecule has one Cl atom, the number of Cl atoms is:

$N = n \times N_A = \frac{0.055}{35.5} \times 6.022 \times 10^{23}$
Compute step by step:
$\frac{0.055}{35.5} = \frac{55}{35500} = \frac{11}{7100} \approx 0.0015493$
Multiply by Avogadro’s number:
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KCET 2024Set B-21 markMCQ
$Q. For which one of the following mixtures is composition uniform throughout? (A) Sand and water (B) Grains and pulses with stone (C) Mixture of oil and water (D) Dilute aqueous solution of sugar$
›Reveal solutionSolution
"Uniform composition throughout" is the definition of a homogeneous mixture (a true solution) — only the sugar solution qualifies.

Step 1 — The concept.

Mixtures are classified by whether their composition is the same at every point:

Homogeneous mixture (solution): solute particles are of molecular/ionic size ( $< 1\,$ nm), uniformly dispersed. Every sample drawn from anywhere has the same composition. Only one phase is visible.

Heterogeneous mixture: two or more distinguishable phases; composition varies from point to point.

Step 2 — Test each option.

(A) Sand and water — sand particles are large and insoluble; they settle to the bottom. Two visible phases → heterogeneous ✗

(B) Grains and pulses with stone — plainly separable solids, each retaining its identity; a scoop from one corner differs from another → heterogeneous ✗ …
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KCET 2022Set B-31 markMCQ
$18g$
›Reveal solutionSolution
Mole fraction is the ratio of moles of one component to total moles. Here, water’s mole fraction is $0.1$ , so the correct option is (C).

The concept here is mole fraction — a way to express concentration in terms of the number of particles (moles) rather than mass. In a mixture, the mole fraction of a component is simply the number of moles of that component divided by the total number of moles of all components. It’s dimensionless and always lies between 0 and 1.

Why does this matter? Because mole fraction directly relates to partial pressures in gases and colligative properties in solutions. For this problem, we just need to convert the given masses into moles using molar masses, then compute the ratio.

Find the moles of water. Water ( $H_2O$ ) has a molar mass of $18 \, \text{g/mol}$ . Given $18 \, \text{g}$ of water:

$n_{\text{water}} = \frac{18}{18} = 1 \, \text{mol}.$

Find the moles of ethyl alcohol. Ethyl alcohol ( $C_2H_5OH$ ) has a molar mass of $46 \, \text{g/mol}$ (carbon: $2 \times 12 = 24$ , hydrogen: $6 \times 1 = 6$ , oxygen: $16$ , total $24+6+16=46$ ). Given $414 \, \text{g}$ of alcohol:

$n_{\text{alcohol}} = \frac{414}{46} = 9 \, \text{mol}.$

Calculate total moles. $n_{\text{total}} = n_{\text{water}} + n_{\text{alcohol}} = 1 + 9 = 10 \, \text{mol}. …$
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KCET 2022Set B-31 markMCQ
$Q. Vacant space in body centered cubic lattice unit cell is about (A) 23% (B) 46% (C) 32% (D) 10%$
›Reveal solutionSolution
Vacant space $= 100\% -$ packing efficiency; for bcc the packing efficiency is $68\%$ , so $32\%$ is empty.

Step 1 — Set up the bcc geometry.

A bcc unit cell has:

$8$ corner atoms, each shared by $8$ cells $\Rightarrow 8\times\tfrac18 = 1$ atom

$1$ atom fully inside at the body centre $\Rightarrow 1$ atom

$Z = 2\ \text{atoms per unit cell}$
Step 2 — Relate radius to edge length.

In bcc the atoms touch along the body diagonal, not along the edge. The body diagonal of a cube of edge $a$ has length $\sqrt3\,a$ , and it contains $4$ radii (corner atom radius + full central atom + corner atom radius):
$\sqrt{3}\,a = 4r \quad\Longrightarrow\quad r = \frac{\sqrt3\,a}{4}$
Step 3 — Compute the packing efficiency.

$\text{P.E.} = \frac{Z \times \tfrac43\pi r^3}{a^3} = \frac{2 \times \tfrac43\pi\left(\tfrac{\sqrt3 a}{4}\right)^3}{a^3}$ …
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Q.What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water?

Concept understanding — Mass Percentage

Mass Percentage: The Intuition

The Precise Definition

Why It Matters in Chemistry

Step-by-Step Example

Why this formula?

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

2. The Natural First Step: The Fraction

3. Why Multiply by 100?

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

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

Method: Molality Formula Using Moles of Solute and Mass of Solvent (in kg)

Steps:

Mistake 1: Confusing Molality with Molarity

Mistake 2: Using the Wrong Mass for the Solvent

Mistake 3: Incorrect Molar Mass Calculation

Q.What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water?

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