Question

A gas sample at stp contains 1.15 g oxygen gas and 1.55 g nitrogen gas.what is the volume of the gas sample?

Answer 1

Explanation:

The given data is as follows.

        Mass of $O_{2}$ = 1.15 g

        Mass of $N_{2}$ = 1.55 g

Therefore, moles of oxygen present will be as follows.

             No. of moles of $O_{2}$ = $\frac{mass}{\text{molar mass}}$

                              = $\frac{1.15 g}{32 g/mol}$

                              = 0.035 mol

             No. of moles of $N_{2}$ = $\frac{mass}{\text{molar mass}}$

                              = $\frac{1.55 g}{28.02 g/mol}$

                              = 0.055 mol

Hence, total no. of moles = moles of $O_{2}$ + moles of $N_{2}$

                              = (0.035 + 0.055) mol

                              = 0.09 mol

Now, it is known that at STP volume is 22.4 L/mol. Hence, volume of the gas sample at STP for 0.09 moles will be as follows.

                    $0.09 mol \times 22.4 L/mol$

                      = 2.01 L

Thus, we can conclude that volume of the given gas sample is 2.01 L.

Answer 2

O2=32 g/ mol
1.15/32=0.035
N2=28 g/mol
1.55/28=0.055
in STP every 22.4 litters is 1 mol