Q89 / 180 · AIIMS 2001 · Chemistry

How many litres of ammonia gas at S.T.P. would be needed to prepare $100 \mathrm{~mL}$ of $2.5 \mathrm{M}$ ammonium hydroxide solution?

  • Correct answer
    A
    $5.6 \mathrm{~L}$
  • B
    $0.056 \mathrm{~L}$
  • C
    $11.2 \mathrm{~L}$
  • D
    $0.56 \mathrm{~L}$

Solution

Molecular weight of $\mathrm{NH}_4 \mathrm{OH}=35$ Now, $35 \mathrm{~g}$ of $\mathrm{NH}_4 \mathrm{OH}$ is needed to prepare $1 \mathrm{M}$ solution in 1 litre $\Rightarrow 35 \mathrm{~g} \mathrm{NH}_4 \mathrm{OH}$ in $1 \mathrm{~L}$ solution $\equiv 1$ molar solution $\Rightarrow 2.5 \times 35 \mathrm{~g} \mathrm{NH}_4 \mathrm{OH}$ in $1 \mathrm{~L}$ solution $\equiv 2.5$ molar solution $$ \begin{aligned} & \Rightarrow \frac{2.5 \times 35}{10} \mathrm{~g} \mathrm{NH}_4 \mathrm{OH} \text { is } 100 \mathrm{~mL} \text { solution } \equiv 2.5 \text { molar solution } \\ & \Rightarrow \quad 8.75 \mathrm{~g} \mathrm{NH}_4 \mathrm{OH} \text { was dissolved. } \\ & \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{NH}_4 \mathrm{OH} \\ & \quad \text { For } 35 \mathrm{~g} \mathrm{NH}_4 \mathrm{OH}, \mathrm{NH}_3 \text { needed }=22.4 \mathrm{~L} \text { at S.T.P } \\ & \Rightarrow \quad \text { for } 8.75 \mathrm{~g} \mathrm{NH}_4 \mathrm{OH}, \mathrm{NH}_3 \text { needed }=\frac{22.4}{35} \times 8.75 \mathrm{~L} \\ &=5.6 \mathrm{~L} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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