Mole Fraction and Molarity JEE 2019: Urea Solution

What is the mole fraction and molarity JEE 2019 question?

The mole fraction and molarity JEE 2019 question requires total solution mass, not water mass, to calculate solution volume. It is a JEE Advanced 2019, Paper 2, Chemistry, Solutions numerical-answer question.

An aqueous solution contains 900 grams of water, and urea accounts for 0.05 of its total moles. Calculate its molarity using the solution density and molar masses below:

ρ_{solution} = 1.2 g c m^{- 3}

M_{urea} = 60 g m o l^{- 1}, M_{water} = 18 g m o l^{- 1}

The question bank classifies it as medium difficulty and gives a 120-second target. Neither is an official exam rating or time limit.

How do you use mole fraction to find the moles of urea?

Use total moles of urea and water in the mole-fraction denominator. The given fraction is not the ratio of urea moles to water moles, so multiplying water moles directly by 0.05 would give the wrong amount.

Let the unknown number of moles of urea be: $x = n_{urea}$

Distinguish this mole amount from the mole-fraction symbol:

x : urea amount in moles, χ_{urea} : dimensionless mole fraction

First calculate the water amount:

n_{water} = \frac{900}{18} = 50 m o l

Apply the definition and follow the supplied solution’s rearrangements:

χ_{urea} = \frac{n_{urea}}{n_{urea} + n_{water}}

0.05 = \frac{x}{x + 50}

0.05 x + 2.5 = x

0.95 x = 2.5

x = \frac{2.5}{0.95} = \frac{50}{19} m o l

Keep this fraction rather than rounding it. That makes the later cancellation visible and avoids carrying an approximate mole amount through the remaining conversions.

How do you convert total solution mass into volume?

Apply the given density to the combined mass of urea and water. Density connects the mass and volume of the same material: solution density requires solution mass, not solvent mass.

Convert the urea amount into mass:

m_{urea} = \frac{50}{19} \times 60 = \frac{3000}{19} g

Then add the water mass:

m_{solution} = 900 + \frac{3000}{19} = \frac{17100}{19} + \frac{3000}{19} = \frac{20100}{19} g

Use the definition of density:

ρ = \frac{m_{solution}}{V_{solution}}, V_{solution} = \frac{m_{solution}}{ρ}

V_{solution} = \frac{20100}{19 \times 1.2} c m^{3}

Molarity needs litres, so convert explicitly:

1000 c m^{3} = 1 L

V_{solution} = \frac{20100}{19 \times 1.2 \times 1000} L = \frac{20.1}{19 \times 1.2} L

Neither water mass nor water volume is the denominator needed for molarity. There is no need to calculate separate volumes for water and urea: the supplied solution density relates their combined mass to the final solution volume.

How do you calculate molarity and report the official answer?

Divide the urea amount by the solution volume in litres. Retain the exact quotient so that the calculation and the supplied official answer remain distinct.

M = \frac{moles of solute}{volume of solution in litres}

Substitute without rounding:

M = \frac{50}{19} \div [\frac{20.1}{19 \times 1.2}]

The common factor cancels:

M = \frac{50}{19} \times \frac{19 \times 1.2}{20.1} = \frac{50 \times 1.2}{20.1} = \frac{60}{20.1} m o l L^{- 1}

Supplied official answer:

2.98 M

This reports the supplied answer, not a claim that the exact quotient conventionally rounds to it. No accepted numerical range is supplied here.

Use this sequence to keep each denominator clear:

mole fraction \to solute moles \to total solution mass \to solution volume \to molarity

Why does using 900 grams as solution mass give the wrong answer?

Using only the water mass underestimates solution volume and therefore overestimates molarity. This is a numerical-answer question, and no wrong options are supplied. The calculation below is an illustrative error, not an exam option.

Keep the correctly calculated urea amount to isolate the volume mistake:

n_{urea} = \frac{50}{19} m o l

If water mass is incorrectly divided by solution density:

V_{wrong} = \frac{900}{1.2} = 750 c m^{3} = 0.750 L

The resulting concentration is:

M_{wrong} = \frac{50 / 19}{0.750} = \frac{200}{57} \approx 3.51 M

The mismatch is specific: 900 grams describes water, while the given density describes the complete solution. Omitting urea mass makes the calculated volume too small; dividing the same mole amount by that smaller volume makes molarity too large.

Before using the following relation, check that mass and density describe the same material: $V = \frac{m}{ρ}$

Write “solution” beside both quantities. This checks the physical meaning rather than relying on a memorised conversion formula.

What two practice questions test the same denominator choices?

Molality uses solvent mass, while mass percentage uses total solution mass. These two author-created practice questions, based on the supplied question, test that distinction. They are not additional verified JEE PYQs.

What is the molality of this urea solution?

The molality is approximately 2.924 moles per kilogram of solvent. Use the mass of water in kilograms, not total solution mass or solution volume.

m_{solvent} = 900 g = 0.900 k g

Molality = \frac{50 / 19}{0.900} = \frac{500}{171} \approx 2.924 m o l k g^{- 1}

Density is unnecessary because this denominator is solvent mass in kilograms. Do not convert the water mass into volume.

What is the mass percentage of urea in this solution?

Urea makes up approximately 14.93 percent of the solution by mass. Divide urea mass by the combined mass of urea and water, then multiply by 100.

Mass percentage = \frac{3000 / 19}{20100 / 19} \times 100 = \frac{10}{67} \times 100 \approx 14.93 %

Compare this with the given mole fraction:

Mole percent = 0.05 \times 100 = 5 %

Five mole percent and approximately 14.93 mass percent are not interchangeable. Urea and water have different molar masses, so their shares of total moles and total mass differ.

Redo both drills with the working covered. Before inserting numbers, write the denominator in words: kilograms of solvent for molality, total solution mass for mass percentage.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

How do you find moles of urea from a mole fraction of 0.05?

The 900 g of water contains 900/18 = 50 mol. If n is the urea amount in moles, its mole fraction gives 0.05 = n/(n + 50), so n = 50/19 mol. Multiplying 50 directly by 0.05 is incorrect because mole fraction uses total solution moles.

What is the answer to the JEE Advanced 2019 urea molarity question?

The supplied official answer is 2.98 M. Using the given data without intermediate rounding gives an exact quotient of 60/20.1 mol/L. The article distinguishes this quotient from the supplied answer and does not provide an accepted numerical range.

Why can't I use 900 g to calculate the solution volume?

The 900 g is water mass, but the given density describes the complete solution. Add the urea mass before dividing by solution density. Using water mass alone gives an incorrectly small volume of 0.750 L and overestimates molarity as approximately 3.51 M.

What is the molality of the urea solution in the JEE 2019 question?

The molality is (50/19)/0.900 = 500/171, or approximately 2.924 mol/kg. Its denominator is the water mass in kilograms, not total solution mass or solution volume. Density is not needed for this calculation.