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A sample of $H_{2} S O_{4}$ (density 1.8 gr/ml) is 90% by weight. What is the volume of the acid that has to be used to make 1 ltr of 0.2 M $H_{2} S O_{4}$ ?

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16 ml

18 ml

12 ml

10 ml

answer is C.

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Detailed Solution

$M = \frac{% [\frac{w}{w}] \times d \times 10}{M w t} = \frac{90 \times 1.8 \times 10}{98} = \frac{90 \times 18}{98} = 16.53$

$M_{1} V_{1} = M_{2} V_{2}$

$16.53 \times V_{1} = 0.2 \times 1$

$V_{1} = \frac{0.2}{16.53} = 0.012 L = 12 m l$

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