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Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Solutions

pH of a 0.1M monobasic acid is found to be 2.Hence its osmotic pressure at a given temp T is

$\begin{array}{1 1}(a)\;0.11RT\\(b)\;0.1RT\\(c)\;0.01RT\\(d)\;1.1RT\end{array}$

1 Answer

Osmotic pressure $(\pi)=iCRT$
Where $C$=Molarity of solution
$\alpha = \large\frac{i-1}{m-1}$
$0.1 = \large\frac{i-1}{2-1}$
i = 1.1
$\therefore \pi= iCRT$
$\therefore = 1.1\times 0.1RT = 0.11RT$
Hence (a) is the correct answer.

 

answered Feb 14, 2014 by sreemathi.v
edited May 27 by sharmaaparna1
What is the use of PH in the question?
 

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