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Concentration Terms - An Introduction (w/w etc)

The density o...

Question

The density of $3 M$ solution of $N a_{2} S_{2} O_{3}$ is $1.25 g m L^{- 1}$ . Calculate, the molalities of $N a^{+}$ and $S_{2} O_{3}^{2 -}$ ions respectively.

7.732 and 3.865

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3.865 and 7.732

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5.643 and 4.321

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None of the above

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Solution

The correct option is A 7.732 and 3.865
Given, $Molarity of N a_{2} S_{2} O_{3} = 3 m o l L^{- 1}$
$∴$ Moles of $N a_{2} S_{2} O_{3} = 3$
$∴ Weight of N a_{2} S_{2} O_{3} = 3 \times 158 = 474 g$
$and Volume of solution = 1 L = 1000 m L$
$∴ Weight of solution = 1000 \times 1.25 = 1250 g$
$∴ Weight of water = 1250 - 474 = 776 g$

$Molality of N a^{+} = \frac{Moles of N a^{+}}{Weight of water (g)} \times 1000 = \frac{6 \times 1000}{776} = 7.732$
Molality of $S_{2} O_{3}^{2 -} = \frac{3 \times 1000}{776} = 3.865$

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