Calculate the molality of a solution containing 20.7 g of potassium carbonate dissolved in 500 mL of solution (assume density of solution $=$ 1 g mL $^{- 1}$ ):

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Solution

Mass of $K_{2} C O_{3} = 20.7 g$

Molar mass of $K_{2} C O_{3} = 138 g m o l^{- 1}$

Moles of $K_{2} C O_{3} = \frac{20.7 g}{138 g m o l^{- 1}}$
$= 0.15 m o l^{- 1}$

Mass of solution $= (500 m L) \times (1 g m L^{- 1})$
$= 500$ g

Amount of water $=$ 500 - 20.7
$=$ 479.3 g Molarity

$= \frac{M o l e s o f s o l u t e}{M a s s o f s o l v e n t i n g r a m} \times 1000$

$= \frac{0.15 m o l}{479.3 g} \times 1000 g k g^{- 1}$

$= 0.313 m o l k g^{- 1} o r = 0.313$

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