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

Twenty grams of hemoglobin in 1L solution at 300K is separated from pure water by a semipermeable.At equilibrium,the height of solution (density $1gcm^{-3}$) in a tube dipped in solution is found to be 83.1mm higher than in the tube dipped in water.If $g=10ms^{-2}$ ,the molar mass of hemoglobin is about

$\begin{array}{1 1}30kgmol^{-1}\\40kgmol^{-1}\\60kgmol^{-1}\\80kgmol^{-1}\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Answer : $60kgmol^{-1}$
The osmotic pressure of solution becomes
$\Pi=h\rho g=(83.1\times 10^{-3}m)(10^3kg m^{-3})(10m s^{-2})=831kg m^{-1}s^{-2}=831Pa$
Using the expression $\Pi=cRT=\large\frac{m/M}{V}$$RT$,we get
$831 Pa=\big(\large\frac{20g}{M}\frac{1}{dm^3}\big)$$(8.314\times 10^3Pa \;dm^{3}K^{-1}mol^{-1})(300K)$
$M=\big(\large\frac{20g\times 8.314\times 10^3\times 300}{831}\big)$$gmol^{-1}=60,000gmol^{-1}=60kgmol^{-1}$
answered Jul 23, 2014 by sreemathi.v
 

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
...