Chemistry - Ideal Gas Question with Solution | TestHub

NTP condition : 273 K and 1 atm pressure

$$\begin{array}{rlr}& \mathrm{PV}=\mathrm{nRT}& \frac{1\times 127.4\times 10^{-3}}{0.0821\times 273}=\mathrm{n}=\text{ number of moles }\end{array}$$

Since mass of gas is collected over water.

Total pressure $=725$ torr

Water vapour pressure $=28$ torr

Partial pressure of gas $=725-25=700$ torr $=700\mathrm{mm}$ of Hg

Temperature $=27^{\circ }\mathrm{C}=300\mathrm{K}$

Using PV = nRT

$\frac{700}{760}\times \mathrm{V}=\frac{1\times 127.4\times 10^{-3}}{0.0821\times 273}\times 0.0821\times 300$

$\mathrm{V}=\frac{76}{70}\times 127.4\times 10^{-3}\times \frac{300}{273}=152\times 10^{-3}$ line

$\mathrm{V}=152{\mathrm{cm}}^3$