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Q.

To convert 12 g of $N a H_{2} P O_{4}$ completely into $N a_{3} P O_{4}$ the volume of 1 molar NaOH required is ? (Molecular mass of $N a H_{2} P O_{4} = 120$

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a

200 $c m^{3}$

b

100 $c m^{3}$

c

80 $c m^{3}$

d

120 $c m^{3}$

answer is A.

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Detailed Solution

$\underset{(N a H_{2} P O_{4})}{\frac{ω}{G E W}} = \underset{(N a O H)}{N . V_{L}}$

$\frac{12}{120 / 2 (n - f a c t o r)} = 1 \times \frac{V_{C C}}{1000}$

$V_{C C} = 200 C C$

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