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For a dilute solution containing a nonvolatile solute,the molar mass of solute evaluated from the osmotic pressure measurement is given as

$\begin{array}{1 1}M_2=\large\frac{m_2}{V}\frac{RT}{\Pi}\\M_2=\large\frac{m_2}{V}\frac{\Pi}{RT}\\M_2=m_2\large\frac{RT}{\Pi}\\M_2=m_2\large\frac{\Pi}{RT}\end{array}$

Answer : $M_2=\large\frac{m_2}{V}\frac{RT}{\Pi}$
We have
$\Pi=cRT=\large\frac{n_2}{V}$$RT=\large\frac{(m_2/M_2)}{V}$$RT$
Hence,$M_2=\large\frac{m_2}{V}\frac{RT}{\Pi}$