MCQ
How many molecules are present in one gram of hydrogen........ $\times {10^{23}}$
- A$6.02$
- ✓$3.01$
- C$2.5$
- D$1.5$
$\therefore $ $1\,gm$ of hydrogen $ = \frac{{6.02 \times {{10}^{23}}}}{2} = 3.01 \times {10^{23}}$ molecule.
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$\mathop {C{H_3} - N{H_2}}\limits_{(P)} \,\,\,\,\,\,\,\,\,\,\,\mathop {Ph - N{H_2}}\limits_{(Q)} $
Reason : Raoult’s law is a special case of Henry’s law.
The above reaction is carried out in a vessel starting with partial pressure $\mathrm{P}_{\mathrm{SO}_{2}}=250\, \mathrm{~m}$ $bar,$ $\mathrm{P}_{0_{2}}=750 \,\mathrm{~m}$ $bar$ and $\mathrm{P}_{\mathrm{SO}_{3}}=0 \,\mathrm{bar}$. When the reaction is complete, the total pressure in the reaction vessel is $.....\mathrm{m}$ $bar.$ (Round off to the Nearest Integer).
