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Volume of Gases and Number of Moles

How many mill...

Question

How many millilitres of $0.1$ N $H_{2} S O_{4}$ solution will be required for complete reaction with a solution containing $0.125$ g of pure $N a_{2} C O_{3}$ ?

A

23.6

mL

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B

25.6

mL

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C

26.3

mL

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D

32.6

mL

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Solution

The correct option is A $23.6$ mL
Number of moles of $N a_{2} C O_{3} = \frac{0.125}{106}$ moles
$n f$ of $N a_{2} C O_{3} = 2$
$n f$ of $H_{2} S O_{4} = 2$
Now, by using
$N_{1} V_{1} = N_{2} V_{2}$
$\Rightarrow M_{1} V_{1} (H_{2} S O_{4}) \times n f_{1} = M_{2} V_{2} (N a_{2} C O_{3}) \times n f_{2}$
as solution of $H_{2} S O_{4}$ is $0.1 M$ .
Thus, $M_{1} \times n_{f_{1}} = M_{1} = 0.1$ as $n = M V$
$\Rightarrow 0.1 \times V_{1} = n_{2} \times n f_{2}$
$\Rightarrow 0.1 \times v_{1} = \frac{0.125}{106} \times 2$
$\Rightarrow V_{1} = 23.58 \times 10^{- 3} L$
$\Rightarrow V_{1} = 23.58 m l$
Thus $23.6$ ml of $H_{2} S O_{4}$ is required .

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