1. Some basic concept of chemistry — Chemistry STD 11 Science — Question

$18\,gm$ ${H_2}O$ =$6.023 \times {10^{23}}$ 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.