What is the molality of the aqueous solution of methyl alcohol ( $C H_{3} O H$ , molar mass $32.0 g m o l^{- 1}$ ) which contains 64 g of methyl alcohol in 200 g of water?

10 mol

k g^{- 1}

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5 mol

k g^{- 1}

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2 mol

k g^{- 1}

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None of these

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Solution

The correct option is A 10 mol $k g^{- 1}$
$M o l a l i t y$ = $\frac{No. of mole of solute}{mass of Solvent}$

Given mass of solute = 64g
Molar mass of methyl alcohol =32
Mole of $C H_{3} O H$ = $\frac{64}{32}$
= 2

mass of solvent in kg =0.2

Hence,

$m$ = $\frac{2}{0.2}$

$= 10 m o l / k g$
answer is A

Suggest Corrections

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(C H_{3} O H, m o l a r m a s s 32.0 g m o l^{- 1})

k g . d m^{- 3}

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(C H_{3} C H_{2} O H, m o l a r m a s s = 46.0 g m o l^{- 1})

C H_{3} C H_{2} O H

= 46.0

g m o l^{- 1}

5.2

C H_{3} O H

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