Some Basic Concepts in Chemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & 4Al + 3{O_2} \to 2A{l_2}{O_3} \cr & {\text{At}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of}}\,Al = 27 \cr & {\text{Thus}}\,4 \times 27g\,{\text{of}}\,Al\,{\text{reacts}}\,{\text{with}}\,{\text{oxygen}} = 3 \times 32g \cr & \therefore 27{\text{g}}\,{\text{of}}\,Al\,{\text{reacts}}\,{\text{with}}\,{\text{oxygen}} = \frac{{3 \times 32}}{{4 \times 27}} \times 27g = 24g \cr} $$

The balanced chemical equation is
\[\underset{\begin{smallmatrix} \\ 24\,g \end{smallmatrix}}{\mathop{Mg}}\,+\underset{16\,g}{\mathop{\frac{1}{2}{{O}_{2}}}}\,\to \underset{\begin{smallmatrix} \\ 40\,g \end{smallmatrix}}{\mathop{MgO}}\,\]
From the above equation, it is clear that,
$$24 g$$  of $$Mg$$  reacts with $$16 g$$  of $${{O_2}}$$.
Thus, $$1.0$$ $$g$$ of $$Mg$$  reacts with
$$\frac{{16}}{{24}}g\,\,{\text{of}}\,{O_2} = 0.67\,g\,{\text{of}}\,{O_2}.$$
But only $$0.56 g$$  of $${{O_2}}$$  is available which is less than $$0.67 g.$$  Thus $${O_2}$$  is the limiting reagent.
Further, $$16$$ $$g$$ of $${{O_2}}$$  reacts with $$24 g$$  of $$Mg.$$
∴ $$0.56 g$$  of $${{O_2}}$$  will react with
$$\eqalign{ & Mg = \frac{{24}}{{16}} \times 0.56 \cr & \,\,\,\,\,\,\,\,\,\, = 0.84\,g \cr} $$
∴ Amount of $$Mg$$  left unreacted
$$\eqalign{ & = \left( {1.0 - 0.84} \right)g\,\,Mg \cr & = 0.16\,g\,Mg \cr} $$

No. of moles of glucose $$ = \frac{{2.82}}{{180}} = 0.01567$$
No. of moles of water $$ = \frac{{30}}{{18}} = 1.667$$
Total no. of moles of solution $$ = 0.01567 + 1.667 = 1.683$$
Mole fraction of glucose $$ = \frac{{0.01567}}{{1.683}} = 0.0093 \approx 0.01$$

$$0.0048 = 4.8 \times {10^{ - 3}}$$

$$\eqalign{ & {\text{Average atomic mass of oxygen}} \cr & = \frac{{\left( {99.763 \times 15.995} \right) + \left( {0.037 \times 16.999} \right) + \left( {0.200 \times 17.999} \right)}}{{100}} \cr & = 15.999\,u \cr} $$

Mass of elements in the body of a healthy human adult are :
Oxygen $$\left( {61.4\% } \right),$$  carbon$$\left( {22.9\% } \right),$$  hydrogen $$\left( {10\% } \right)$$  and nitrogen $$\left( {2.6\% } \right).$$
Weight of the person $$ = 75\,kg$$
Mass due to $$^1H = 75 \times \frac{{10}}{{100}} = 7.5\,kg$$
On replacing $$^1H$$  by $$^2H,7.5\,kg$$    mass would replace with $$15\,kg.$$
∴ Net mass gained by person $$ = \left( {15 - 7.5} \right)kg = 7.5\,kg$$

$$\eqalign{ & {\text{Given chemical}}\,e{q^n} \cr & {M_2}C{O_3} + 2HCl \to 2MCl + {H_2}O + C{O_2} \cr & 1gm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.01186\,mol \cr & {\text{From the above chemical}}\,e{q^n}. \cr & n{M_2}C{O_3} = nC{O_2} \cr & \frac{1}{{{\text{Molar mass of}}\,\,{M_2}C{O_3}}} = 0.01186 \cr & \therefore \,\,{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{M_2}C{O_3} = \frac{1}{{0.01186}} \cr & M = 84.3\,gm/mol \cr} $$

Key Concept Mole is the biggest unit to measure number of molecules/atoms/ions.
$$\because \,\,1\,mole\,$$  of water contains molecules $$ = 6.02 \times {10^{23}}$$
$$\therefore 18\,moles$$   of water contain molecules
$$ = 18 \times 6.02 \times {10^{23}}\,molecules$$
Now, $$1 mole$$  of water $$= 18 g$$  of water
Hence, number of molecules in $$18 g$$  of water
$$ = 6.02 \times {10^{23}}$$
and $$1.8 g$$  of water contains $$ = 6.02 \times {10^{22}}molecules$$

$$\eqalign{ & {\text{Let mole fraction of}}\,{O_2}\,{\text{is}}\,x \cr & 40 = 32 \times x + 80\left( {1 - x} \right) \cr & {\text{or}}\,\,x = \frac{5}{6} \cr & a:b = x\,:\left( {1 - x} \right) = \frac{5}{6}:\frac{1}{6} \cr & {\text{When ratio is changed}} \cr & {M_{mixture}} = 32 \times \frac{1}{{6}} + 80 \times \frac{5}{6} = 72 \cr} $$

Standard molar volume is the volume occupied by $$1\,mole$$  of a gas at $$STP.$$
$$0.1784\,g$$   of $$He$$  occupies volume $$ = 1\,L$$
$$4\,g\left( {1\,mole} \right)$$   of $$He$$  occupies $$\frac{4}{{0.1784}} = 22.4\,L$$