The unit of ebullioscopic constant is _______________. (i) $K\ kg\ mol^{-1}$ or $K\ (molality)^{-1}$ (ii) $mol\ kg\ K^{-1}$ or $K^{-1}(molality)$ (iii) $kg\ mol^{-1}\ K^{-1}$ or $K^{-1}(molality)^{-1}$ (iv) $K\ mol\ kg^{-1}$ or $K\ (molality)$

The unit of ebullioscopic constant is ________. (i) Kkgmol⁻¹ or K(molality)⁻¹ (ii) molkgK⁻¹ or K⁻¹(molality) (

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Q. The unit of ebullioscopic constant is _______________.

(i)

K\ kg\ mol^{-1}

K\ (molality)^{-1}

(ii)

mol\ kg\ K^{-1}

K^{-1}(molality)

(iii)

kg\ mol^{-1}\ K^{-1}

K^{-1}(molality)^{-1}

(iv)

K\ mol\ kg^{-1}

K\ (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.

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

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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The ebullioscopic constant $K_b$ relates boiling point elevation to molality, so its unit must be temperature per molality — giving $K\ kg\ mol^{-1}$ or $K\ (molality)^{-1}$ , which matches option (i).

The ebullioscopic constant (also called the boiling point elevation constant) appears in the formula for boiling point elevation:

\Delta T_b = K_b \cdot m

Here, $\Delta T_b$ is the elevation in boiling point (measured in Kelvin, K), and $m$ is the molality of the solution (measured in mol/kg, i.e., moles of solute per kilogram of solvent). The constant $K_b$ is a property of the solvent, not the solute.

To find the unit of $K_b$ , we simply rearrange the equation:

K_b = \frac{\Delta T_b}{m}

Now substitute the units:

$\Delta T_b$ has unit: K (Kelvin)
$m$ has unit: mol/kg (or mol kg $^{-1}$ )

So:

\text{Unit of } K_b = \frac{\text{K}}{\text{mol kg}^{-1}} = \text{K} \cdot \text{kg mol}^{-1}

That is: $K\ kg\ mol^{-1}$ .

Why not the other options?

Option (ii) gives $mol\ kg\ K^{-1}$ — that would be the reciprocal of $K_b$ , i.e., $1/K_b$ .

Option (iii) gives $kg\ mol^{-1}\ K^{-1}$ — that has an extra $K^{-1}$ in the denominator, which would make $K_b$ dimensionless when multiplied by molality, which is wrong.

Option (iv) gives $K\ mol\ kg^{-1}$ — that is $K \cdot$ (molality), which is actually the unit of $\Delta T_b$ , not $K_b$ .
A quick check using the formula

If $K_b$ had units of $K\ kg\ mol^{-1}$ , then $K_b \times m$ gives: …

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Common Mistakes on the Unit of Ebullioscopic Constant ( $K_b$ )

The Correct Answer

The ebullioscopic constant $K_b$ has units of $K\ kg\ mol^{-1}$ or equivalently $K\ (molality)^{-1}$ .

So option (i) is correct.

Mistake #1: Confusing $K_b$ with its reciprocal or cryoscopic constant

What students do wrong:

They pick option (iii) — $kg\ mol^{-1}\ K^{-1}$ — thinking $K_b$ has $K^{-1}$ in the unit.

Why this happens:

Students mix up the ebullioscopic constant ( $K_b$ ) with the cryoscopic constant ( $K_f$ ) or with the reciprocal relationship in formulas like $\Delta T_b = K_b \cdot m$ .

How to avoid:

Remember the formula:

\Delta T_b = K_b \cdot m

Here $\Delta T_b$ is in K and $m$ (molality) is in mol/kg.

Solve for $K_b$ :

K_b = \frac{\Delta T_b}{m}

So units are:

\frac{K}{mol/kg} = K \cdot \frac{kg}{mol} = K\ kg\ mol^{-1}

Key insight: $K_b$ tells you how many Kelvin the boiling point rises per unit molality — so Kelvin is in the numerator, not denominator.

Mistake #2: Reversing the numerator and denominator

What students do wrong:

They pick option (iv) — $K\ mol\ kg^{-1}$ — which has mol in the numerator and kg in the denominator.

Why this happens:

Students misremember the formula as $\Delta T_b = \frac{K_b}{m}$ or confuse molality ( $mol/kg$ ) with its reciprocal ( $kg/mol$ ).

How to avoid:

Write the formula clearly:

\Delta T_b = K_b \times m

Molality $m$ has units $mol\ kg^{-1}$ .
For the product to give Kelvin, $K_b$ must cancel the $mol\ kg^{-1}$ part:

K_b \times (mol\ kg^{-1}) = K \quad \Rightarrow \quad K_b = K \times kg\ mol^{-1}

Trick: The unit of $K_b$ is the reciprocal of the unit of molality.

Mistake #3: Picking option (ii) — $mol\ kg\ K^{-1}$

What students do wrong:

They think $K_b$ has mol in the numerator and Kelvin in the denominator.

Why this happens: …

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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.The unit of ebullioscopic constant is _______________.

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)

Concept: Ebullioscopic Constant (Boiling Point Elevation Constant)

Formula to remember

Method: Dimensional Analysis

Steps

Common Mistakes on the Unit of Ebullioscopic Constant ( $K_b$ )

The Correct Answer

Mistake #1: Confusing $K_b$ with its reciprocal or cryoscopic constant

Mistake #2: Reversing the numerator and denominator

Mistake #3: Picking option (ii) — $mol\ kg\ K^{-1}$

Q.The unit of ebullioscopic constant is _______________.

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