JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Some Basic Concepts of Chemistry

250 g solution of D-glucose in water contains

10.8%

of carbon by weight. The molality of the solution is nearest to
(Given: Atomic Weights are H, 1u ; C, 12u ; O, 16u)

Answer: (B)

Solution

D-glucose has the formula

C X_{6} H X_{12} O X_{6}

with molar mass 180 g/mol. The mass fraction of carbon in glucose is

\frac{72}{180} = 0.4

.

Given that the solution contains 10.8% carbon by weight:

Mass of C = \frac{10.8 \times 250}{100} = 27 g

Using the carbon fraction in glucose:

Mass of glucose = \frac{27}{0.4} = 67.5 g

Moles of glucose:

Moles = \frac{67.5}{180} = 0.375 mol

Mass of solvent (water):

250 - 67.5 = 182.5 g = 0.1825 kg

Molality:

m = \frac{0.375}{0.1825} \approx 2.06 mol/kg

The answer is (B) 2.06.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option