Calculate the molarity and molality of a $13 %$ solution (by weight of sulphuric acid). Its density is $1.090$ g/ml.

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Solution

13 % solution of sulphuric acid contains 13 g $H_{2} S O_{4}$ in 100 g of solution.

Weight of water $= 100 - 13 = 87 g$

Volume of $H_{2} S O_{4}$ solution $= \frac{w e i g h t o f s o l u t i o n}{d e n s i t y} = \frac{100 g}{1.090 g / m l} = 91.7 m l = \frac{91.7 L}{1000} = 0.0917 L$

Molarity of $H_{2} S O_{4}$ solution $= \frac{m o l e o f H_{2} S O_{4}}{V o l u m e o f s o l u t i o n} = \frac{\frac{13}{98}}{0.0917} = 1.45 M$

Molality of $H_{2} S O_{4}$ solution $= \frac{m o l e o f H_{2} S O_{4}}{w e i g h t o f w a t e r i n k g} = \frac{\frac{13}{98}}{\frac{87}{1000}} = 1.525 m$

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