Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in 75g of benzene.

Molar mass of C2H4O2:12×2+1×4+16×2=60gmol−1Moles of C2H4O2=2.5g60gmol−1 = 0.0417molMass of banzene in Kg = 75g/1000gkg−1=75×10−3kg Molality of C2H4O2 molesof C2H4O2kg of benzene =0.0417mol×1000gkg−175g=0.556molkg−1

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75g of benzene.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Molar mass of $C_{2} H_{4} O_{2} :$
$12 \times 2 + 1 \times 4 + 16 \times 2 = 60 g {mol}^{- 1}$
Moles of $C_{2} H_{4} O_{2} = \frac{2.5 g}{60 g {mol}^{- 1}}$ = 0.0417mol
Mass of banzene in Kg = $75 g / 1000 g {kg}^{- 1}$
$= 75 \times 10^{- 3} kg$
$Molality of C_{2} H_{4} O_{2} \frac{molesof C_{2} H_{4} O_{2}}{kg of benzene}$
$= \frac{0.0417 mol \times 1000 g {kg}^{- 1}}{75 g} = 0.556 mol {kg}^{- 1}$