Molality of $18$ M $H_{2} S O_{4}$ (d = $1.8$ $g m L^{- 1}$ ) is :

36 m o l k g^{- 1}

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200 m o l k g^{- 1}

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500 m o l k g^{- 1}

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18 m o l k g^{- 1}

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Solution

The correct option is C $500 m o l k g^{- 1}$
The expression for the molality M is $\frac{1000 M}{(1000 d - M m)}$
Substitute values in the above expression.
$M = \frac{1000 \times 18}{(1000 \times 1.8 - 18 \times 98)} = 500 m o l k g^{- 1}$

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