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Molarity

How many mL o...

Question

How many $m L$ of $H_{2} S O_{4}$ of density $1.8 g m L^{- 1}$ containing $92.5 %$ by volume of $H_{2} S O_{4}$ should be added to $1 L$ of $40 %$ solution of $H_{2} S O_{4}$ (density $1.3 g m L^{- 1}$ ) in order to prepare $50 %$ solution of $H_{2} S O_{4}$ (density $1.4 g m L^{- 1}$ )?
186

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Solution

The correct option is A 186
Molarity of solution containing $92.5 % H_{2} S O_{4}$ may be calculated as:
Molarity ( $M_{1}$ ) = volume of $H_{2} S O_{4} \times D e n s i t y \times \frac{1000}{98} \times 100$
= $92.5 \times 1.8 \times \frac{1000}{98} \times 100$
= $16.99 M$
Molarity of solution containing $40$ ,
$M_{2} = 40 \times 1.3 \times \frac{1000}{98} \times 100$
= $5.31 M$

Molarity of required solution containing $50 % H_{2} S O_{4}$ :
$M = 50 \times 1.4 \times \frac{1000}{98} \times 100$
= $7.14 M$
Let $V$ litre of a solution with molarity ( $M_{1}$ ) is added to $1$ litre of solution with molarity ( $M_{2}$ )
To prepare $(1 + V)$ litre of solution with molarity $M$ , then:
$M_{1} \times V_{1} + M_{2} \times V_{2}$ = $M \times V$
$16.99 \times V + 5.31 \times 1$ = $7.14 \times (1 + V)$
= $9.85 V = 1.83$
= $V = \frac{1.83}{9.85}$ = $0.186 L$
Or $186 m L$

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