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

$$\eqalign{ & N = \frac{{{N_1}{V_1} + {N_2}{V_2} + {N_3}{V_3}}}{{{\text{Total volume}}}} \cr & = \frac{{50 \times 10 + 25 \times 12 + 40 \times 5}}{{1000}} \cr & = 1N \cr} $$

$$\eqalign{ & {\text{Normality of ferrous }}amm.{\text{ sulphate}} \cr & {\text{ = }}\frac{{3.92 \times 1000}}{{392 \times 100}} \cr & = 0.1\,\left( {Eq.\,wt\,{\text{of}}\,FAS\,{\text{is}}\,{\text{392}}} \right) \cr & {N_1}{V_1} = {N_2}{V_2} \cr & 20 \times 0.1 = 18 \times {N_2}\,\,\,\,\,\,\,\,\,{N_2} = 0.111 \cr & 1\,g\,ev.\,{\text{of}}\,KMn{O_4} = 31.6\,g \cr & 0.111\,g\,ev.\,{\text{of}}\,KMn{O_4} = 31.6 \times 0.111 = 3.5\,g. \cr} $$

No. of nitrogen atoms $$ = \frac{{{\text{Mass}}\,{\text{in}}\,{\text{grams}}}}{{{\text{Atomic}}\,{\text{wt}}{\text{.}}}} = \frac{{28}}{{14}} = 2$$
No. of oxygen atoms $$ = \frac{{{\text{Mass}}\,{\text{in}}\,{\text{grams}}}}{{{\text{Atomic}}\,{\text{wt}}{\text{.}}}} = \frac{{80}}{{16}} = 5$$
∴ Formula of compound is $${N_2}{O_5}.$$

$$\eqalign{ & {\text{Density}} = \frac{{{\text{Mass}}}}{{{\text{Volume}}}} \cr & \therefore \,\,{\text{Mass = Density }} \times {\text{ Volume}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.12\,g\,m{L^{ - 1}} \times 1.5\,mL \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.68\,g \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.7\,g \cr} $$

$$\eqalign{ & 4Fe + 3{O_2} \to 2F{e_2}{O_3} \cr & {\text{Each }}1{\text{ }}g{\text{ of }}Fe{\text{ requires }}0.428{\text{ }}g{\text{ of}}\,\,{O_2}. \cr & 2Mg + {O_2} \to 2MgO \cr & {\text{Each }}1{\text{ }}g{\text{ of }}Mg{\text{ requires }}0.67{\text{ }}g{\text{ of }}{O_2}. \cr & {P_4} + 5{O_2} \to {P_4}{O_{10}} \cr & {\text{Each }}1{\text{ }}g{\text{ of }}{P_4}{\text{ requires }}1.29{\text{ }}g{\text{ of }}{O_2}. \cr & {C_3}{H_8} + 5{O_2} \to 3{O_2} + 4{H_2}O \cr & {\text{Each }}1{\text{ }}g{\text{ of }}{C_3}{H_8}\,{\text{requires }}3.64{\text{ }}g{\text{ of }}{O_2}. \cr} $$

$$\begin{array}{l}3D+4E\stackrel{80\%}{\to }5C+A\\ 9mole14mole\frac{5}{3}\times 9\times 0.8=12mole\\ \underset{12mole}{3C}+\underset{4mole}{5G}\stackrel{50\%}{\to }6B+F\\ \text{Limiting Reagent is }G\\ \therefore \text{Moles of }B\text{ formed}=\frac{6}{5}\times 4\times 0.5=2.4\end{array}$$

$${K_2}C{r_2}{O_7}$$   and $$KMn{O_4}$$   both are good oxidizing agents in acidic medium. The reactions are as follows :
$$\eqalign{ & C{r_2}{O_7}^{2 - } + 14{H^ + } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O \cr & Mn{O_4}^{2 - } + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O \cr & {\text{Normality of}}\,\,C{r_2}{O_7}^{2 - } \cr & = {\text{Molarity }} \times {\text{ no}}{\text{. of electrons involved}} \cr & = 0.04 \times 6 \cr & = 0.24\,N \cr & {\text{Normality of}}\,\,MnO_4^ - \cr & {\text{ = Molarity }} \times {\text{ no}}{\text{. of electrons involved}} \cr & = 0.03 \times 5 \cr & = 0.15\,N \cr & \mathop {{N_1}{V_1}}\limits_{\left( {{K_2}C{r_2}{O_7}} \right)} = \mathop {{N_2}{V_2}}\limits_{\left( {KMn{O_4}} \right)} \cr & \Rightarrow 0.24 \times 50 = 0.15 \times {V_2} \cr & \Rightarrow {V_2} = \frac{{0.24 \times 50}}{{0.15}} = 80\,c{m^3} \cr} $$

$$\mathop {{C_3}{H_8}}\limits_{22.4\,L} + \mathop {5{O_2}}\limits_{5 \times 22.4\,L} \to 3C{O_2} + 4{H_2}O$$
For the combustion of $$22.4L$$  propane, oxygen required $$ = 5 \times 22.4\,L$$
For the combustion of $$1 L$$  of propane
oxygen required $$ = \frac{{5 \times 22.4}}{{22.4}}L = 5L$$

$$\eqalign{ & 18\,g\,\,{\text{of}}\,\,{H_2}O = 1\,mol \cr & 180\,g\,\,{\text{of}}\,\,{H_2}O = 10\,mol \cr} $$
$$18\,g\,\,{\text{of}}\,\,{H_2}O$$    $$ = 6.023 \times {10^{23}}\,\,{\text{molecules}}\,\,{\text{of}}\,\,{H_2}O$$
$$\eqalign{ & 180\,g\,\,{\text{of}}\,\,{H_2}O = \frac{{6.023 \times {{10}^{23}}}}{{18}} \times 180 \cr & = 6.023 \times {10^{24}}\,\,{\text{molecules}}\,\,{\text{of}}\,\,{H_2}O \cr} $$

TIPS/Formulae:
From the molarity equation.
$$\eqalign{ & {M_1}{V_1} + {M_2}{V_2} = MV \cr & {\text{Let}}\,\,M\,\,{\text{be}}\,{\text{the}}\,{\text{molarity}}\,{\text{of}}\,{\text{final}}\,{\text{mixture,}} \cr & M = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{V}\,{\text{where}}\,V = {V_1} + {V_2} \cr & M = \frac{{480 \times 1.5 + 520 \times 1.2}}{{480 + 520}} = 1.344\,M \cr} $$