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Stoichiometric Calculations

To convert 12...

Question

To convert 12 g of ${N a H}_{2} {P O}_{4}$ complete to ${N a}_{3} {P O}_{4}$ , volume of 1 M NaOH solution required is:

200 ml

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220 ml

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500 ml

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300 ml

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Solution

The correct option is A 200 ml
$N a H_{2} P O_{4} (s) + 2 N a O H (a q) \to N a_{3} P O_{4} (a q) + 2 H_{2} O (l)$

Moles of $N a H_{2} P O_{4} = \frac{12}{120} = 0.1 m o l e s$

0.1 moles of $N a H_{2} P O_{4}$ requires 0.2 moles of $N a O H$ .
Given,
1 mole $N a O H$ is present in 1000 ml of solution.

Therefore, 0.2 moles of $N a O H$ $⟶$ $0.2 \times 1000$ .

Hence, the volume required = $200 m l$ .

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