Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
0 votes

An antifreeze solution is prepared from 222.6 of ethylene glycol and 200g of water.Calculate the molality of the solution.

$\begin{array}{1 1}17.95m\\16.75m\\14.75m\\12.75m\end{array} $

Can you answer this question?

1 Answer

0 votes
Given :
Mass of solution (ethylene glycol)=222.6g
Molar mass of ethylene glycol $(C_2H_4(OH)_2)=62$
Therefore moles of ethylene glycol =$\large\frac{222.6}{62}=$$3.59$ mole
Mass of water (solvent) =200g
Molality =$\large\frac{\text{Moles of solute}}{\text{Mass of solvent (gm)}}$$\times 1000$
$\Rightarrow \large\frac{3.59}{200}$$\times 1000$
$\Rightarrow 17.95m$
answered Aug 7, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App