Question

A container is partitioned by a semipermeable membrane (SPM), open from the top as shown in the figure. Initially, the left partition has 1 mole of $NaCl$ (non-volatile) in 90 g of water while the right partition has 1 mole of glucose (non-volatile) in 72 g of water at room temperature. After a long time, when steady state is reached by the transfer of some liquid water from the one partition to the other, which of the following options is true?

• A. 18 g water is transferred from $NaCl$ solution to glucose solution
• B. 18 g water is transferred from glucose solution to $NaCl$ solution
• C. 9 g water is transferred from $NaCl$ solution to glucose solution
• D. 9 g water is transferred from glucose solution to $NaCl$ solution

Answer 1

The correct option is B 18 g water is transferred from glucose solution to $NaCl$ solution

$\Rightarrow$ Steady state will reach when osmotic pressure of both solutions will become same.

${\pi }_{left}={\pi }_{right}$
${\pi }_{left}=i{C}_{left}RT$
${\pi }_{right}=i{C}_{right}RT$

Let $x$ gm of solvent (water) transferred from left to right.
Therefore,
$i{C}_{left}RT=i{C}_{right}RT$
$\Rightarrow 2\times \frac{1}{90-x}=\frac{1}{72+x}$
(where $i=2$ for $NaCl$ and $i=1$ for glucose)
$\Rightarrow 144+2x=90-x$
$\Rightarrow x=-18$
We assumed that water has been transferred from left to right. Since we are getting negative value of $x$ thus 18 g of water transferred from right (glucose) to left $\left(NaCl\right)$.