The density of 3 M solution of ${Na}_{2} S_{2} O_{3}$ is 1.25 g m $L^{- 1}$ . Select the correct statements :

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The % of weight of ${Na}_{2} S_{2} O_{3}$ is 37.92

The mole fraction of ${Na}_{2} S_{2} O_{3}$ is 0.065

The molality of ${Na}^{+}$ is 8.732

The molality of $S_{2} O_{3}^{2-}$ is 3.866

answer is A, B, D.

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Detailed Solution

Given, $M_{{Na}_{2} S_{2} O_{2}} = 3$

$∴ {Moles of Na}_{2} S_{2} O_{3} = 3$

${Wt. of Na}_{2} S_{2} O_{3} 3 × 158 = 474 g$

$Wt. of solution = 1000 × 1.25 = 1250g$

Volume of solution = 1000 mL

$∴$ Wt. of water = 1250 – 474 = 776g

Now % by weight of ${Na}_{2} S_{2} O_{3} = \frac{{wt. of Na}_{2} S_{2} O_{3}}{wt. of solution} × 100 = \frac{474}{1250} × 100 = 37.92$

${Mole fraction of Na}_{2} S_{2} O_{3} = \frac{{mole of Na}_{2} S_{2} O_{3}}{{moles of Na}_{2} S_{2} O_{3} {+ moles of H}_{2} O} = \frac{3}{3+ (\frac{776}{18})} = 0.065$ ${Molality of Na}^{+} = \frac{{moles of Na}^{+}}{wt. of water (in kg)} = \frac{6}{776/1000} = 7.732 m$