MCQ
The largest number of molecules is in
- ✓$34\,g$ of water
- B$28\,g$ of $C{O_2}$
- C$46\,g$ of $C{H_3}OH$
- D$54\,g$ of ${N_2}{O_5}$
✓
Answer
Correct option: A.
$34\,g$ of water
a
$(a)$ $(a)$ $34\,gm$ of water
$(a)$ $(a)$ $34\,gm$ of water
$18\,gm$ ${H_2}O$ =$6.023 \times {10^{23}}$ molecule $\therefore $ $34\,gm$ ${H_2}O$ = $\frac{{6.023 \times {{10}^{23}}}}{{18}} \times 34$ $ = 11.37 \times {10^{23}}$ mole
$(b)$ $28\,gm$ of $C{O_2}$ $44\,gm$ $C{O_2}$$ = 6 \times {10^{23}}$ molecules
$28\,gm$ $C{O_2}$ $ = \frac{{6 \times {{10}^{23}}}}{{44}} \times 28 = 3.8 \times {10^{23}}$
$(c)$ $46\,gm$ of $C{H_3}OH$ $32\,gm$ $C{H_3}OH\; = 6 \times {10^{23}}$ molecules
$46\,gm$ $C{H_3}OH\; = \frac{{6 \times {{10}^{23}}}}{{32}} \times 46 = 8.625 \times {10^{23}}$
$(d)$ $108\,gm$ of ${N_2}{O_5} = 6 \times {10^{23}}$ molecules
$54\,gm$ of ${N_2}{O_5}\; = \frac{{6 \times {{10}^{23}}}}{{108}} \times 54 = 3 \times {10^{23}}$ molecules.
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