Solution Found!

Solved: An ethylene glycol solution contains 21.2 g of ethylene glycol (C2H6O2) in 85.4

Chapter 13, Problem 106P

(choose chapter or problem)

Get Unlimited Answers
QUESTION:

An ethylene glycol solution contains 21.2 g of ethylene glycol \(\mathrm {(C_2H_6O_2)}\) in 85.4 mL of water. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

Questions & Answers

QUESTION:

An ethylene glycol solution contains 21.2 g of ethylene glycol \(\mathrm {(C_2H_6O_2)}\) in 85.4 mL of water. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

ANSWER:

 

 

Step 1 of 3

Here, we are going to calculate the boiling point and freezing point of the solution .

 Calculation of the amount of ethylene glycol in mol

Mass of water = Density x volume = 1.0 g/L x 85.4 L = 85.4 g

We know,

Molar mass of ethylene glycol  = 62.07 g/mol

Mass of ethylene glycol  = 21.2 g

Amount of glucose in mol = 21.2 g/62.07 g/mol = 0.3415 mol

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back