Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Questions  >>  JEEMAIN and NEET  >>  Chemistry  >>  Solutions
Answer
Comment
Share
Q)

At room temperature $25^{\large\circ}C$,the expression for total pressure of mixture of methanol and ethanol solution is given by P(in torr)$=135+120X$ where $X$ is mol fraction of methanol,Hence

$\begin{array}{1 1}(a)\;\text{vapour pressure of pure ethanol in 135 torr}\\(b)\;\text{vapour pressure of pure ethanol in 255 torr}\\(c)\;\text{both (a) and (b) correct}\\(d)\;\text{mixture is completely immiscible}\end{array}$

1 Answer

Comment
A)
Given $P=135+120X$
If $X=1$,then vapour pressure of methanol $P=P^0$
$\therefore P=135+120(1)$
$\Rightarrow 135+120=255torr$
If $X=0$,then vapour pressure of methanol $P=P^0$
$\therefore P=135+120(0)$
$\Rightarrow P=135$
Hence (c) is the correct answer.
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.
...