Molarity (M) of an aqueous solution containing x g of anhyd. ? in 500 mL solution at 32°C is M. Its molality will be _____ m

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Solution

Q.
Molarity (M) of an aqueous solution containing $x$ g of anhyd. ${CuSO}_{4}$ in 500 mL solution at 32°C is $2 \times 10^{- 1}$ M. Its molality will be _____ $\times 10^{- 3}$ m. (nearest integer).

[Given density of the solution = 1.25 g/mL] [2024]

Ans.
(164)

   $W_{solution} = d_{solution} \times V_{solution} = 1.25 \frac{g}{mL} \times 500 mL = 625 g$

   $n_{solute} = {Molarity}_{solution} \times V_{solution} = 2 \times 10^{- 1} M \times 0.5 L = 0.1 mol$

$W_{solute} = n_{solute} \times molar mass of solute = 0.1 mol \times 159.5 {gmol}^{- 1} = 15.95 g$

   $W_{solvent} = W_{solution} - W_{solute} = 625 g - 15.95 g = 609.05 g$

   $m = \frac{n_{solute} \times 1000}{W_{solvent} (in g)} = \frac{0.1 \times 1000}{609.05} m = 0.1642 m = 164 \times 10^{- 3} m$

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