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

$$\eqalign{ & \,\,\,\,\,\,\,3.0223 \cr & + \,\,\,\,\,\,5.041 \cr & \,\,\,\,\,\,\,\overline {8.0633} \cr} $$
Since, 5.041 has only 3 digits after the decimal point, the result should be reported to 3 digits after decimal point.

In alkaline medium, $$KMn{O_4}$$  is reduced to $${K_2}Mn{O_4}$$
$$\eqalign{ & KI + {H_2}O \to KOH + HI \cr & \mathop {2KMn{O_4}}\limits^{ + 7} + 2KOH \to 2{K_2}\mathop {\mathop {Mn{O_4}}\limits^{ + 6} + {H_2}O + \left[ O \right]}\limits^{} \cr} $$
Hence, one mole of $$KMn{O_4}$$  is reduced by one mole of $$KI.$$

In $$\left( {PbO,Pb{O_2}} \right)$$    oxygen $$\left( {16g,32g} \right)$$   combine with a fixed mass of lead $$\left( {Pb} \right),$$  bearing a simple ratio, 16 : 32 or 1 : 2 ( Law of multiple proportions ).

$$\eqalign{ & {M_1}{V_1} = {M_2}{V_2} \cr & 5\,M \times 500\,mL = {M_2} \times 1500\,mL \cr & \therefore \,\,{M_2} = \frac{{5 \times 500}}{{1500}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.66\,M \cr} $$

$$\eqalign{ & \mathop {CuO}\limits_{79.5\,g} + \mathop {{H_2}}\limits_{22.4\,L} \to Cu + {H_2}O \cr & 22.4\,L\,\,{\text{of}}\,\,{H_2} \equiv 79.5\,g\,\,{\text{of}}\,\,CuO \cr} $$
$$2.80\,\,L\,\,{\text{of}}\,\,{H_2} \equiv \frac{{79.5}}{{22.4}} \times 2.80$$       $$ = 9.9\,g\,\,{\text{of}}\,\,CuO$$

$$\eqalign{ & {\text{(A) }}18\,g\,{\text{of}}\,{H_2}O = 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{H_2}O \cr & \therefore 36\,g\,{\text{of }}{H_2}O = 2 \times 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{H_2}O \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}12.04 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{H_2}O \cr & {\text{(B)}}\,\,{\text{28}}\,{\text{g}}\,{\text{of}}\,CO = 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,CO \cr & {\text{(C)}}\,\,46\,g{\text{ of}}\,{C_2}{H_5}OH = 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{C_2}{H_5}OH \cr & ({\text{D}})\,\,108\,\,g\,\,{\text{of}}\,{N_2}{O_5} = 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{N_2}{O_5} \cr & \therefore 54\,g\,\,{\text{of}}\,{N_2}{O_5} = \frac{1}{2} \times 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{N_2}{O_5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.01 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{N_2}{O_5} \cr & \therefore 36\,g\,{\text{of}}\,{\text{water}}\,{\text{has}}\,{\text{highest}}\,{\text{number}}\,{\text{of}}\,{\text{molecules}}{\text{.}} \cr} $$

\[\begin{align} & \text{Decomposition of}\,A{{g}_{2}}C{{O}_{3}}\,\text{gives silver} \\ & \underset{\begin{smallmatrix} 2\times 276\,g \\ 2.76\,g \end{smallmatrix}}{\mathop{2A{{g}_{2}}C{{O}_{3}}}}\,\to \underset{\begin{smallmatrix} 4\times 108\,g \\ 2.16\,g \end{smallmatrix}}{\mathop{4Ag}}\,+2C{{O}_{2}}+{{O}_{2}} \\ \end{align}\]

$$\eqalign{ & {M_1}{V_1} + {M_2}{V_2} = {M_3}{V_3}\left( {{V_3} = {V_1} + {V_2}} \right) \cr & 1.5 \times 480 + 1.2 \times 520 = {M_3} \times 1000 \cr & {M_3} = 1.34\,M \cr} $$

$$\eqalign{ & 2KMn{O_4}\,\, + \,\,3{H_2}S{O_4}\,\,\, \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5\left[ O \right] \cr & \left[ {MnO_4^ - + \,8{H^ + } + \,5{e^ - }\,\,\, \to M{n^{2 + }} + 4{H_2}O} \right] \times 2\left[ {SO_3^{2 - } + {H_2}O \to SO_4^{2 - } + 2{H^ + } + 2{e^ - }} \right] \times 5 \cr & \overline {2MnO_4^ - + 6{H^ + } + 5SO_3^{2 - } \to 2M{n^{2 + }} + 5SO_4^{2 - } + 3{H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr} $$

$$5\,moles$$   of sulphite ions react with $$ = 2\,moles$$   of $$MnO_4^ - $$
So, $$1{\text{ }}mole$$   of sulphite ions react with $$ = \frac{2}{5}moles$$   of $$MnO_4^ - .$$

$$\eqalign{ & 2S{O_2} + \,\,\,\,\,\,{O_2} \to \,\,\,2S{O_3} \cr & 10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & 10 - 2x\,\,\,\,\,15 - x\,\,\,\,\,\,2x \cr & \because \,\,2x = 8\,\,\,x = 4 \cr & {\text{Hence, remaining,}}\,\,S{O_2} = 10 - 8 = 2\,moles, \cr & {O_2} = 15 - 4 = 11\,moles \cr} $$