The molality of a 10 % ( v / V ) solution of di-bromine solution in CCl _ 4 (carbon tetrachloride) is x '. x = _____ × 10 ^ - 2 M . (Nearest integer) [Given : molar mass of Br _ 2 = 160g mol ^ - 1 atomic mass of C = 12g

Correct answer: 0 10 % v / Vmeans10ml solute in 90ml solvent. mass of solute( Br _ 2 ) = 10 × 3 . 2 = 32g mass of solvent( CCl _ 4 ) = 90 × 1 . 6g molality(m) = (massofsolute)/(molarmassofsolute) × (1000)/(massofsolvent) m = (32 × 1000)/(160 × 90 × 1 . 6) = 1 . 388 m = 138 . 8 × 10 ^ - 2 = 139

JEE Main 20231 FebruaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 1 February
Shift: Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2023 · 1 February · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

The molality of a 10 % ( v / V ) solution of di-bromine solution in...

The molality of a $10 % ( v / V )$ solution of di-bromine solution in ${CCl}_{4}$ (carbon tetrachloride) is $x$ '. $x = _____ \times {10}^{- 2}M$ . (Nearest integer)

[Given : molar mass of ${Br}_{2} = 160g{mol}^{- 1}$

atomic mass of $C = 12g{mol}^{- 1}$

atomic mass of $Cl = 35 . 5g{mol}^{- 1}$

density of dibromine $= 3 . 2g{cm}^{- 3}$

density of {CCl}_{4} = 1 . 6g{cm}^{- 3}]$

Correct answer

Step-by-step solution

$10 % v / V\text{means}10ml$ solute in $90ml$ solvent.

mass of solute\left({Br}_{2}\right) = 10 \times 3 . 2 = 32g $</p><p>mass of solvent\left({CCl}_{4}\right) = 90 \times 1 . 6g$

molality(m) = $\dfrac{massofsolute}{molarmassofsolute} \times \dfrac{1000}{massofsolvent}$

$m = \dfrac{32 \times 1000}{160 \times 90 \times 1 . 6} = 1 . 388$

$m = 138 . 8 \times {10}^{- 2} = 139$

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