MCQ
The number of molecules in $4.25 \,g$ of ammonia are
- A$0.5 \times {10^{23}}$
- ✓$1.5 \times {10^{23}}$
- C$3.5 \times {10^{23}}$
- D$1.8 \times {10^{32}}$
According to the mole concept
$17\,gm$ $N{H_3}$ has molecules $ = 6.02 \times {10^{23}}$
$\therefore \,\,1\,gm$ $N{H_3}$ has molecules $ = \frac{{6.02 \times {{10}^{23}}}}{{17}}$
$\therefore $ $4.25\,gm$ $N{H_3}$ has molecules $ = \frac{{6.02 \times {{10}^{23}} \times 4.25}}{{17}} = 1.5 \times {10^{23}}\,molecule$
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