What volume of $H_{2} S O_{4}$ (density of of 1.85 g $c m^{- 3}$ ) and what mass of wate must be taken to prepare $100 c m^{3}$ of 15% soln of sulphuric acid (den-1.10 g $c m^{- 3})$ ?

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Solution

Volume of the solution $= 100 c m^{2}$
Density of the solution $= 1.10 g / c m^{3}$
Therefore, mass of $100 c m^{3}$ of solution $= 100 \times 1.10 g = 110 g$
The given solutions is 15%. Therefore, 100g of solution contains 15g of $H_{2} S O_{4}$
Then, mass $H_{2} S O_{4}$ in 10g (100 $c m^{3}$ ) of solution $= \frac{15 g}{100 g} \times 110 g = 16.5 g$
Mass of water in 110g (100 $c m^{3}$ ) of solution = (110 - 16.5)g = 93.5g
To obtain $100 c m^{3}$ of 15% solution acid, we require the given information,
Mass of water = 93.5 g
Mass of $H_{2} S O_{4}$ (100% pure) = 16.5g
Since, the given sulphuric acid is 95% pure, hence,
Mass of $H_{2} S O_{4}$ (95%\) will be required =
$\frac{100 \times 16.5 g}{95} = 17.37 g$
Density of 95% $H_{2} S O_{4} = 1.85 c m^{- 3}$

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