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Electrolytes

Solid NaHCO...

Question

Solid $N a H C O_{3}$ will be neutralised by 40.0 mL of 0.1 M $H_{2} S O_{4}$ solution. What would be the weight of solid $N a H C O_{3}$ in gram?

A

0.672 g

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B

6.07 g

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C

17 g

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D

20 g

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Solution

The correct option is C 0.672 g
The required neutralisation reaction is
$2 N a H C O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O + 2 C O_{2}$
Mole 2 mole 1 mole
ratio = 168 g =98 g
m-moles of $H_{2} S O_{4} = M \times V_{m L} = 40.0 \times 0.1 = 4 m - m o l e$
or m-mol of $N a H C O_{3}$ neutralised $= 4 \times 2 = 8 m - m o l e$
But, $m - m o l e = \frac{w}{m} \times 1000$
$8 = \frac{w}{84} \times 1000$
$\Rightarrow w = \frac{84 \times 8}{1000} \Rightarrow w = 0.672 g$

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