MCQ
A gaseous mixture contains oxygen and nitrogen in the ratio of $1: 4$ by weight. Therefore, the ratio of their number of molecules is
- A$7: 32$
- B$1: 8$
- C$3: 16$
- D$1: 4$
✓
Answer
A. $7: 32$
Explanation:
Let mass of oxygen $=1 g$, Then mass of nitrogen $=4 g$
Mol. wt. of $N ^2=28 g, Mol$. wt. of $O ^2=32 g$
28 g of $^2$ has $=6.02 \times 10^{23}$ molecules of nitrogen
4 g of $N ^2$ has $=\frac{6.02 \times 10^{23}}{28} \times 4$ molecules of nitrogen
$=\frac{6.02 \times 10^{23}}{7}$ molecules of nitrogen
32 g of $O ^2$ has $=6.02 \times 10^{23}$
$\therefore 1 g$ of O $^2$ has $=\frac{6.02 \times 10^{23}}{32} \times 1$
$=\frac{6.02 \times 10^{23}}{32}$ molecules of oxygen
Explanation:
Let mass of oxygen $=1 g$, Then mass of nitrogen $=4 g$
Mol. wt. of $N ^2=28 g, Mol$. wt. of $O ^2=32 g$
28 g of $^2$ has $=6.02 \times 10^{23}$ molecules of nitrogen
4 g of $N ^2$ has $=\frac{6.02 \times 10^{23}}{28} \times 4$ molecules of nitrogen
$=\frac{6.02 \times 10^{23}}{7}$ molecules of nitrogen
32 g of $O ^2$ has $=6.02 \times 10^{23}$
$\therefore 1 g$ of O $^2$ has $=\frac{6.02 \times 10^{23}}{32} \times 1$
$=\frac{6.02 \times 10^{23}}{32}$ molecules of oxygen
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