Battery acid ia $4.22 M$ aqueous $H_{2} S O_{4}$ solution, and has density fo $1.21 g c m^{- 3}$ . What is the molality of $H_{2} S O_{4}$ ? H=1, S=32, O=16
( $m = 5.298 m o l k g^{- 1}$ )

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Solution

The molarity of the solution is 4.22M. Therefore 4.22 mol of $H_{2} S O_{4}$ is present in 1000 $c m^{3}$ of solution.
Now M(molar mass of $H_{2} S O_{4}$ )=98 $g (m o l)^{- 1}$
W(mass of $H_{2} S O_{4}) = 4.22 \times 98 = 413.56 g$
mass of the solution= $d e n s i t y \times v o l u m e = 1.21 \times 1000 = 1210 g$
mass of water(solvent)= $1210 g - 413.56 g = 796.44 g = 0.79644 k g$
$m (m o l a l i t y) = \frac{n_{s o l u t e}}{mass of solvent(in kg)}$
$m = \frac{4.22 m o l}{0.79644 k g} = 5.298 m o l (k g)^{- 1}$

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