Question

3.55 grams of O2 is enclosed in a 1.5-L container. How many additional grams of O2 should be added to the container to increase the volume to 3.0 L at the same temperature and pressure?

Answer 1

Volume of gas $(V)\propto$ Number of moles of gas $(n) \propto$ Mass of gas enclosed(M)

So,$\frac{V_1}{V_2}=\frac{M_1}{M_2}$

$V_1=1.5L,M_1=3.55\ g;V_2=3\ L,M_2=?$

$\frac{1.5}{3}=\frac{3.55}{M_2}$ $\implies M_2=3.55\times 2=7.10\ g$

Additional gas to be added$=7.10\ g-3.55\ g=3.55\ g$