Ask Questions, Get Answers

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

A 0.1539 molal aqueous solution of cane sugar (mol mass =$342gmol^{-1}$) has a freezing point of 271K while the freezing point of pure water is $273.15K$.What will be the freezing point of an aqueous solution containing 5g of glucose (mol mass =$180gmol^{-1}$) per 100g of solution.

$\begin{array}{1 1}269.27K\\367.29K\\169.27K\\469.27K\end{array} $

Can you answer this question?

1 Answer

0 votes
Molality =0.1539m
$\Delta T_f=273.15-271K$
$\Delta T_f=K_f.m$
$m=\large\frac{\Delta T_f}{K_f}$
$\Delta T_f=K_f\times m$
$\Rightarrow \large\frac{2.15}{0.1539}\times \frac{5}{180}\times \frac{1000}{95}$
$\Rightarrow 4.08K$
Freezing point of solution =273.15-4.08K
$\Rightarrow 269.27K$
answered Aug 8, 2014 by sreemathi.v

Related questions

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