Chemistry - Mole Concept Question with Solution | TestHub

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 $$\begin{array}{rl}& \Rightarrow \frac{2.5\times 35}{10}\mathrm{g}{\mathrm{NH}}_4\mathrm{OH}\text{ is }100\mathrm{mL}\text{ solution }\equiv 2.5\text{ molar solution }\\ & \Rightarrow 8.75\mathrm{g}{\mathrm{NH}}_4\mathrm{OH}\text{ was dissolved. }\\ & {\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NH}}_4\mathrm{OH}\\ & \text{ For }35\mathrm{g}{\mathrm{NH}}_4\mathrm{OH},{\mathrm{NH}}_3\text{ needed }=22.4\mathrm{L}\text{ at S.T.P }\\ & \Rightarrow \text{ for }8.75\mathrm{g}{\mathrm{NH}}_4\mathrm{OH},{\mathrm{NH}}_3\text{ needed }=\frac{22.4}{35}\times 8.75\mathrm{L}\\ & =5.6\mathrm{L}\end{array}$$