Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Questions  >>  CBSE XII  >>  Chemistry  >>  Solutions
Answer
Comment
Share
Q)

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} $

1 Answer

Comment
A)
Molality =0.1539m
$\Delta T_f=273.15-271K$
$\qquad=2.15K$
$\Delta T_f=K_f.m$
$m=\large\frac{\Delta T_f}{K_f}$
$0.1539=\large\frac{2.15}{K_f}$
$K_f=\large\frac{2.15}{0.1539}$
$W_2=5g$
$M_2=180gmol^{-1}$
$\Delta T_f=K_f\times m$
$\Rightarrow \large\frac{2.15}{0.1539}\times \frac{5}{180}\times \frac{1000}{95}$
$\Rightarrow 4.08K$
Freezing point of solution =273.15-4.08K
$\Rightarrow 269.27K$
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...