Q. The mole percent of oxygen gas present in gaseous mixture containing 14.0 gm of nitrogen and 32.0 gm of oxygen is:
(a) 50. (b) 33.3. (c)66.6. (d) 40.

Open in App

Solution

Given mass of nitrogen = 14g
Molar mass of nitrogen gas = 28 g
Given mass of oxygen = 32g
Molar mass of oxygen gas = 32g
No of moles of oxygen in the mixture, $X_{1}$ = $\frac{m a s s o f o x y g e n}{m o l a r m a s s o f o x y g e n} = \frac{32}{32} = 1$
No of moles of Nitrogen in the mixture, $X_{2}$ = $\frac{m a s s o f n i t r o g e n}{(m o l e m a s s o f n i t r o g e n} = \frac{14}{28} = 0.5$
Mole percent of oxygen = $\frac{X_{1}}{X_{1} + X_{2}} \times 100 = \frac{1}{1 + 0.5} \times 100 = \frac{100}{1.5} = 66.66$ %

Suggest Corrections

Similar questions

Q. The total number of moles present in a gaseous mixture containing

8

g of oxygen,

11.2

L of

C O_{2}

and

3 \times

10^{22}

molecules of

N_{2}

at STP is:

Q. A cylinder of compressed gas contains nitrogen and oxygen in the ratio

3 : 1

by mole. If the cylinder is known to contain

2.5 \times 10^{4} g

of oxygen, what is the total mass of the gas mixture?

Q. A gaseous mixture contains oxygen and nitrogen in

1 : 4

ratio by weight. The ratio of their number of molecules is:

The total number of moles present in a gaseous mixture containing $8 g$ of oxygen, $11.2 l i t r e s$ of $C O_{2}$ and $3 \times 10^{22}$ molecules of $N_{2}$ at STP is

Q. A gaseous mixture contains oxygen and nitrogen in the ratio of

1 : 4

by weight. The ratio of their number of moleules is :

Join BYJU'S Learning Program

Q. The mole percent of oxygen gas present in gaseous mixture containing 14.0 gm of nitrogen and 32.0 gm of oxygen is:(a) 50. (b) 33.3. (c)66.6. (d) 40.

Q. The mole percent of oxygen gas present in gaseous mixture containing 14.0 gm of nitrogen and 32.0 gm of oxygen is:
(a) 50. (b) 33.3. (c)66.6. (d) 40.