Q29 / 90 · JEE Main 2019 (09 Apr Shift 2) · Chemistry

What would be the molality of 20% (mass/mass) aqueous solution of $K I$ ? (molar mass of $K I = 166 g m o l^{- 1}$ )

A
$1.35$
Correct answer
B
$1.51$
C
$1.08$
D
$1.48$

Solution

Molality

= \frac{M o l e s o f s o l u t e}{w e i g h t o f s o l v e n t (g)} \times 1000

100 g m

of solution contains

20 g m K I

So, the moles of

K I

= \frac{20}{166}

Weight of solvent

= 100 - 20 = 80 g m

Molality

= \frac{20 \times 1000}{166 \times 80}

= 1.51 M o l a l

Hence, the correct answer is option: 'B'.

Relevant Questions

The molarity of a (mass mass) aqueous solution of a monobasic acid is __________ M(Nearest integer) [Given: Density of aqueous solution of (28 Jan, 2025 (M)Concentration terms
Concentrated nitric acid is labelled as by mass. The volume in mL of the solution which contains 30 g of nitric acid is ____. Given : Densit28 Jan, 2025 (E)Concentration terms
The hydrocarbon with molar mass and carbon has degree of unsaturation.24 Jan, 2025 (E)Concentration terms
Density of 3 M NaCl solution is . The molality of the solution is :22 Jan, 2025 (E)Concentration terms
20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is _______ . (Nearest integer)22 Jan, 2025 (E)Concentration terms
Molarity \(( {M})\) of an aqueous solution containing \(x {~g}\) of anhyd. \( {CuSO}_4\) in \(500 {~mL}\) solution at \(32{ }^{ } {C}\) is \09 Apr, 2024 (M)Concentration terms