Skip to content
dronacharya.ai
← JEE Advanced 2016 · Paper 1
ChemistryInteger+3Question 32 of 54

JEE Advanced 2016 · Paper 1, Chemistry question 32

The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm−32.0\ \mathrm{g\,cm^{-3}}. The ratio of the molecular weights of the solute and solvent, (MWsoluteMWsolvent)\left(\dfrac{MW_{solute}}{MW_{solvent}}\right), is

Answer and worked solution

Official answer: 9

Molarity = molality. With n2n_2 mol of solute, n2/V(L)=n2/W1(kg)n_2/V(\text{L})=n_2/W_1(\text{kg}). So the volume in mL equals the solvent mass in g.

Density 2.0 g cm⁻³. The solution's mass is 2V=2W12V=2W_1, so the solute mass W2=W1W_2=W_1.

Mole fraction 0.1.
n2n1=0.10.9=19  ⟹  W2/M2W1/M1=M1M2=19.\frac{n_2}{n_1}=\frac{0.1}{0.9}=\frac19\implies\frac{W_2/M_2}{W_1/M_1}=\frac{M_1}{M_2}=\frac19.

Answer: 9

Topics:Concentration terms
← Previous questionNext question →