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Molality

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Question

Calculate the molality of a $1 L$ solution of $68 % H_{2} S O_{4} (\frac{w}{v}),$ if the density of the solution is given as $1.80 g m L^{- 1}$ .

A

6.8 m

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B

6.2 m

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C

6.94 m

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D

10.8 m

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Solution

The correct option is B $6.2 m$
68% $(\frac{w}{v}) H_{2} S O_{4} = 68 g H_{2} S O_{4}$ in $100 m L$ solution.
i.e., $680 g$ $H_{2} S O_{4}$ in $1000 m L$ solution.
Mass of solution $=$ volume $\times$ density
$= 1000 m L \times 1.80 g / m L$
$= 1800 g$
Mass of solvent = $1800 - 680 = 1120 g$
Again, Molar mass of $H_{2} S O_{4} = 2 \times 1 + 32 + 16 \times 4 = 98 g$
Number of moles of solute,
$\frac{Mass}{Molar mass} = \frac{680}{98} = 6.94$ mol
Molality:
$= \frac{No. of moles of solute}{Mass of solvent in kg} = \frac{(6.94 \times 1000)}{1120} = 6.2 m$

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