150g of acetic acid was contaminated with 10 . 2g ascorbic acid ( C _ 6 H _ 8 O _ 6 ) to lower down its freezing point by (x × 10 ^ - 1 ) ° C . The value of x is____ (Nearest integer). [Given K _ f = 3 . 9Kkg mol ^ - 1

Correct answer: 0 Given 150gofsolvent CH _ 3 COOH And as solute 10 . 2g ascorbic acid = (10 . 2)/(176) = 0 . 058 moles Δ T _ f = (x × 10 ^ - 1 ) ° C Δ T _ f = k _ f · molality = 3 . 9 × (0 . 058)/(150) × 1000 = 1 . 5 ° C = 15 × 10 ^ - 1 ° C So, value of x=15.

JEE Main 202228 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 28 July
Shift: Morning Shift
Subject: Chemistry
Chapter: Solutions

JEE Main 2022 · 28 July · Morning Shift · Chemistry · Solutions

150g of acetic acid was contaminated with 10 . 2g ascorbic acid ( C...

150g

of acetic acid was contaminated with

10 . 2g

ascorbic acid \left({C}_{6}{H}_{8}{O}_{6}\right)

to lower down its freezing point by \left(x \times {10}^{- 1}\right) ° C

. The value of

x

is____ (Nearest integer). [Given

{K}_{f} = 3 . 9Kkg{mol}^{- 1}

; Molar mass of ascorbic acid

= 176g{mol}^{- 1}

]

Correct answer

Step-by-step solution

Given $150gofsolvent{CH}_{3}COOH$
And as solute $10 . 2g$ ascorbic acid $= \dfrac{10 . 2}{176} = 0 . 058$ moles
${\Delta T}_{f} = \left(x \times {10}^{- 1}\right) ° C$
${\Delta T}_{f} = {k}_{f} \cdot$ molality
$= 3 . 9 \times \dfrac{0 . 058}{150} \times 1000$
$= 1 . 5 ° C$
$= 15 \times {10}^{- 1} ° C$
So, value of x=15.

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