Molarity of $H_{2} S O_{4}$ is $18$ M. If density d is $8$ g $c m^{- 3}$ , calculate its concentration in mol $k g^{- 1}$ solvent $(H_{2} O)$ , i.e., molality.

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Solution

Molarity is defined as number of moles of solute dissolved in per litre of solution.
Molarity, $M = \frac{n}{V}$
Given tha M=18 M and density of solution $ρ = 8 g / m L$
$∴$ mass of 1 mL solution = 8g
or mass of 1L (1000mL) solution $= 1000 \times 8 = 8000 g$
In 1000mL solution, number of moles of $H_{2} S O_{4}$ is:
$n = M o l a r i t y \times V o l u m e (L) = 18 \times 1 = 18$ mol
mass of solution=mass of solvent + mass of solute
or, $m_{s o l u t i o n} = m_{s o l v e n t}$ +Molar mass of $H_{2} S O_{4} \times$ number of moles
upon substitution we get
$8000 = m_{s o l v e n t} + 98 \times 18$
$m_{s o l v e n t} = 8000 - 1764 = 6236 g = 6.236 k g$
molality is number of moles of solute present in per kg of solvent
$m = n / m_{s o l v e n t} = 18 / 6.236 = 2.886$ molal

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