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Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2024 — 8 April, Shift 2 — Question 83

Molality of an aqueous solution of urea is 4.44 m4.44\ m. Mole fraction of urea in solution is X×10−3X \times 10^{-3}. The value of XX is _____\_\_\_\_\_ (integer answer).

Answer: 74

Numerical answer — enter this value.

Step-by-step solution

According to molality definition, we have

4.44 m=4.44 mol urea in 1 kg water\mathrm{4.44\ m = 4.44\ mol\ urea\ in\ 1\ kg\ water}

Moles of water =100018=55.56 mol= \mathrm{\frac{1000}{18} = 55.56\ mol}

Mole fraction of urea, x=4.444.44+55.56\mathrm{x = \frac{4.44}{4.44 + 55.56}}

x=4.4460=0.074\mathrm{x = \frac{4.44}{60} = 0.074} x=74×10−3\mathrm{x = 74 \times 10^{-3}}

Thus, the value of XX is 7474.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 2
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic
Concentration Terms and Their Interconversion

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