logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
0 votes

One mole each of the following solutes are taken in 1000g of water

 

$\begin{array}{1 1}(A)\;KCl\\(B)\;K_2SO_4\\(C)\;Na_3PO_4\\(D)\;glucose\end{array}$

Assuming 100% ionisation of the electrolyte,relative decrease in vapour pressure will be in order

$\begin{array}{1 1}(a)\;A > B > C > D\\(b)\;A > C > B > D\\(c)\;A > B > D > A\\(d)\;C > B > A > D\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Relative decrease in vapour pressure is directly proportional to molality and Van't Hoff factor (i)
$Na_3PO_4 \longrightarrow 3 Na^+ + PO_4^{3-}$. Thus, 'i' for $Na_3PO_4$ is 4
$K_2SO_4 \longrightarrow 2K^{+} + SO_4^{2-}$. Thus, 'i' for $K_2SO_4$=3,
$KCl \longrightarrow K^+ +Cl^-$. Thus 'i' for $KCl =2$
Glucose doesn't dissociate in water. So, $Glucose(s) \longrightarrow Glucose(l)$. Thus, $i=1$
Since, particle conventration decrease with increase in vapour pressure
$Na_3PO_4 > K_2SO_4 > KCl > glucose$
$C > B > A > D$
Hence (d) is the correct answer.
answered Feb 14, 2014 by sreemathi.v
edited Jul 20, 2014 by mosymeow_1
 

Related questions

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