The density of 3 M sodium thiosulphate is $1.25 g m L^{- 1}$ Identify the correct statement among the following.

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% by weight of sodium thiosulphate is 37.92.

All of the above.

The mole fraction of sodium thiosulphate is 0.065

The molality of $N a^{+}$ is 7.74 and $S_{2} O_{3}^{2 -}$ is 3.87

answer is D.

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Detailed Solution

$M = \frac{M a s s % \times d \times 10}{M_{2}} 3 = \frac{M a s s % \times 1.25 \times 10}{158} [∵ M_{N a_{2} S_{2} O_{3}} = 158 g m o l^{- 1}] M a s s % = \frac{3 \times 158}{1.25 \times 10} = 37.92$
Also, $\frac{1}{m} = \frac{d}{M} - \frac{M_{2}}{1000}; \frac{1}{m} = \frac{1.25}{3} - \frac{158}{1000}$
$\Rightarrow m = \frac{3000}{776} = 3.87$
Thus, molality of $N a^{+}$ ions = 2 × 3.87 = 7.74 m
and molality of $S_{2} O_{3}^{2 -}$ ions = 3.87 m
Moles of $N a_{2} S_{2} O_{3}$ in 1000 g of water = 3.87
Moles of solvent $= \frac{1000}{18} = 55.55$
Mole fraction of $N a_{2} S_{2} O_{3} = \frac{n_{2}}{n_{1} + n_{2}} = \frac{3.87}{3.87 + 55.55}$
$= \frac{3.87}{59.42} = 0.065$