Q.A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
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🔒 Start your 14-day free trial to unlock the full solution →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 25050=0.2 of the whole. Multiply by 100 to get the percentage: 0.2×100=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
Mass percentage of component=Total mass of mixtureMass of that component×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.
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.
Total mass=25 g (glucose)+175 g (water)=200 g
Step 3: Apply the formula.
Mass percentage of glucose=20025×100%=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:
Fraction of component=Total mass of mixtureMass of component
For the apple example:
500 g100 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:
Percentage=(Fraction)×100
So:
0.2×100=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:
Mass percentage=Total mass of mixtureMass of component×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: …
Take a convenient 1 kg (1000 g) sample of the water solution.
- Percent by mass. 15 ppm means 15 parts per million by mass, i.e. 15 g of chloroform per 106 g of solution.
% by mass=10615×100=1.5×10−3 %
- Molality. In 1000 g of solution the mass of chloroform is
so the sample holds 0.015 g of CHCl3, and the mass of water is essentially 1000 g=1 kg. M(CHCl3)=12+1+3(35.5)=119.5 g mol−1 …
15 ppm=1000 g of solution0.015 g
Working on a clean 1 kg basis, 15 ppm chloroform gives 1.5×10−3 % by mass and a molality of 1.25×10−4 m.
Step 1 — Read what "ppm by mass" means.
Parts per million by mass tells us the mass of solute present in every 106 mass units of solution. So 15 ppm means
15 ppm=106 g of solution15 g of CHCl3
Step 2 — (i) Convert to percent by mass.
Percent by mass is parts per hundred, so multiply the mass fraction by 100:
% by mass=10615×100=1.5×10−3 %
Step 3 — Choose a convenient sample size for molality.
Molality needs moles of solute per kilogram of solvent. Take 1 kg=1000 g of the solution. Scaling the ppm ratio down to this 1000 g sample:
mass of CHCl3=10615×1000 g=0.015 g
Since the chloroform is only a trace contaminant, the mass of water (the solvent) is essentially the whole sample:
mass of water≈1000 g=1 kg
Step 4 — Moles of chloroform.
The molar mass of CHCl3 is …
Method: Parts Per Million (ppm) to Mass Percentage and Molality Conversion
This problem uses two key concepts: ppm as a ratio and molality definition.
Step 1 — Understand what 15 ppm means
15 ppm (by mass) means:
15 grams of chloroform per 106 grams of solution.
So we write:
Mass of CHCl3=15 g
Mass of solution=106 g
Step 2 — Express as percent by mass
Percent by mass is:
Percent by mass=mass of solutionmass of solute×100
Substitute:
Percent by mass=10615×100=1.5×10−3%
Answer (i): 1.5×10−3%
Step 3 — Find molar mass of chloroform (CHCl3)
Atomic masses:
- C = 12
- H = 1
- Cl = 35.5
MCHCl3=12+1+3(35.5)=12+1+106.5=119.5 g/mol
Step 4 — Calculate molality
Molality (m) is:
m=kg of solventmoles of solute
Moles of chloroform:
moles=119.515≈0.1255 mol
Mass of solvent (water):
Since the solution mass is 106 g and solute is 15 g: …
Here are the common mistakes students make when solving this Mass Percentage and Molality problem, along with clear strategies to avoid each.
Mistake 1: Confusing ppm with a direct percentage
The Error:
Students often think 15 ppm means 15% or they try to convert by simply moving the decimal (e.g., writing 0.15% or 1.5%).
Why it happens:
They don’t internalise that ppm = parts per million = mass of solutionmass of solute×106. Percent is per hundred, so the conversion factor is 104 (since 106/102=104).
How to avoid:
Always write the definition first:
ppm=mass of solutionmass of solute×106
Then convert to percent:
percent by mass=104ppm
So for 15 ppm:
percent=10415=1.5×10−3%
Key check: 15 ppm is a tiny amount — your answer should be a very small percentage (not 0.15% or 15%).
Mistake 2: Using the wrong molar mass for chloroform (CHCl3)
The Error:
Students mis-count atoms — e.g., forgetting there are 3 chlorine atoms, or using atomic mass of carbon as 12.0 instead of 12.01 (though for this problem, 12 is acceptable if the exam allows rounding).
Why it happens:
Rushing through the formula without careful counting.
How to avoid:
Write the atomic masses clearly:
- C = 12.01 g/mol
- H = 1.008 g/mol
- Cl = 35.45 g/mol
Then sum:
MCHCl3=12.01+1.008+3(35.45)=12.01+1.008+106.35=119.368 g/mol
Round to 119.4 g/mol for most exam purposes.
Mistake 3: Assuming 15 ppm means 15 g of solute in 106 g of water (instead of solution)
The Error:
Students take the solvent mass as exactly 106 g and ignore the solute mass when calculating molality.
Why it happens:
They confuse ppm by mass (solute/solution) with a ratio involving only the solvent.
How to avoid:
Remember:
- ppm by mass = mass solutionmass solute×106
- Molality = mass of solvent (in kg)moles solute
For dilute solutions (like 15 ppm), the mass of solute is negligible compared to solvent, so you can approximate:
Mass of solution ≈ mass of solvent
But always state this approximation in your solution.
Mistake 4: Forgetting to convert solvent mass to kilograms for molality
The Error:
Using grams instead of kg in the denominator of molality.
Why it happens:
Molality is defined as moles per kg of solvent, but students plug in grams.
How to avoid:
Write the formula every time:
molality=mass of solvent (kg)moles of solute
If you have mass in grams, divide by 1000.
Mistake 5: Incorrect unit handling in the molality calculation
The Error: …
- COMEDK 2026Set 2026-M1 markMCQQ.An aqueous solution of an unknown solute " X " is prepared by adding 4.0 g of it into 2.0 moles of water. What is the mass percent of " X " in the aqueous solution? (A) 20 (B) 40 (C) 15 (D) 10
›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:
mass percent=mass of solutionmass of solute×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 g/mol.
mass of water=2.0 mol×18.0 molg=36 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).
total mass=4.0 g+36 g=40 g
- Calculate the mass percent
- KCET 2025Set D-41 markMCQQ.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
xA=nA+nBnA=molmol
Moles divided by moles ⇒ unitless (and it always lies between 0 and 1).
- Mass percent (W/W)
%(w/w)=mass of solutionmass of solute×100=gg×100
Grams divided by grams ⇒ unitless (the "%" is a pure number, not a unit).
- Molality
m=mass of solvent in kgmoles of solute=molkg−1
Moles divided by mass — two different kinds of quantity ⇒ has units.
- Molarity
M=volume of solution in Lmoles of solute=molL−1
Moles divided by volume ⇒ 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 (molkg−1) — fails.
- (C) Molality has units — fails. …
- COMEDK 2025Set 2025-M1 markMCQQ.If X is a haloalkane with a single Chlorine atom per molecule and the percentage of Cl is 55 , what would be the number of Cl atoms present in 0.1 g of the haloalkane? Atomic mass of Cl=35.5 g/mol (A) 6.022×1022 (B) 1.2044×1021 (C) 9.328×1020 (D) 9.329×1023
›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 9.328×1020 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:
M35.5×100%=55%
So:
M35.5=0.55
Solving:
M=0.5535.5=64.545… g/mol
(We can keep it as 0.5535.5 for now.)
- Find moles of haloalkane in 0.1 g Moles of compound:
n=Mmass=35.5/0.550.1=35.50.1×0.55
Simplify:
n=35.50.055 mol
- Number of molecules (and thus Cl atoms) Since each molecule has one Cl atom, the number of Cl atoms is:
N=n×NA=35.50.055×6.022×1023
Compute step by step:
35.50.055=3550055=710011≈0.0015493
Multiply by Avogadro’s number:
- KCET 2024Set B-21 markMCQQ.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 (<1nm), 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 ✗ …
- KCET 2022Set B-31 markMCQQ.An aqueous solution of alcohol contains 18g of water and 414g of ethyl alcohol. The mole fraction of water is (A) 0.7 (B) 0.9 (C) 0.1 (D) 0.4
›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 (H2O) has a molar mass of 18g/mol. Given 18g of water:
nwater=1818=1mol.
- Find the moles of ethyl alcohol. Ethyl alcohol (C2H5OH) has a molar mass of 46g/mol (carbon: 2×12=24, hydrogen: 6×1=6, oxygen: 16, total 24+6+16=46). Given 414g of alcohol:
nalcohol=46414=9mol.
- Calculate total moles. ntotal=nwater+nalcohol=1+9=10mol.…
- KCET 2022Set B-31 markMCQQ.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 ⇒8×81=1 atom
- 1 atom fully inside at the body centre ⇒1 atom
Z=2 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 3a, and it contains 4 radii (corner atom radius + full central atom + corner atom radius):
3a=4r⟹r=43a
Step 3 — Compute the packing efficiency.
P.E.=a3Z×34πr3=a32×34π(43a)3 …
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