Calculate the molar fraction of ethyleneglycol(C2H6O2)in a solution containing 20%by mass. 

Answers (1)

 

20% of C2H6O2 by mass is present.

That means solution has 20 g of ethylene glycol and 80 g of water.

Now,

Molar mass of C2H6O2 = 12 x 2 + 1 x 6 + 2 x 16 = 62 g mol-1 .

Moles of C2H6O2 = 20 / 62 = 0.322 moles

Moles of water = 80 / 18 = 4.444 moles

Mole fraction of ethylene glycol = 0.322 / 0.322 + 4.444 = 0.068

Mole fraction of water = 1 - 0.068 = 0.932

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions