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

$$1\,mole\,CC{l_4}\,{\text{vapours}} = 12 + {\text{ }}4 \times 35.5$$        $$ = 154g$$
$${\text{At }}STP{\text{, volume of 1 }}mole{\text{ of a gas}}$$       $$ = 22.4L$$
$$\eqalign{ & {\text{Thus, }}154\,g = 22.4\,L \cr & \therefore \,{\text{Density of }}CC{l_4}\,{\text{vapours = }}\frac{{154}}{{22.4}}g\,{L^{ - 1}} \cr & = 6.87\,g\,{L^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Since molarity of}}\,{\text{solution is }}3.60{\text{ }}M{\text{. It means }}3.6{\text{ }}moles \cr & {\text{of}}\,{H_2}S{O_4}\,{\text{is present in its 1 litre solution}}{\text{.}} \cr & {\text{Mass of }}3.6{\text{ moles}}\,{\text{of }}{H_2}S{O_4} \cr & = {\text{Moles }} \times {\text{ Molecular mass = }}3.6{\text{ }} \times {\text{ }}98{\text{ }}g{\text{ }} = {\text{ }}352.8{\text{ }}g \cr & \therefore \,1000{\text{ }}ml{\text{ solution has }}352.8{\text{ }}g{\text{ of}}\,{H_2}S{O_4} \cr & {\text{Given that }}29{\text{ }}g{\text{ of}}\,{H_2}S{O_4}\,{\text{is present in}} = 100{\text{ }}g{\text{ of}}\,{\text{solution}} \cr & \therefore \,352.8{\text{ }}g{\text{ of}}\,{H_2}S{O_4}\,{\text{is present in}} \cr & = \frac{{100}}{{29}} \times 352.8\,g{\text{ of solution}} = 1216{\text{ }}g{\text{ of solution}} \cr & {\text{Density}} = \frac{{{\text{Mass}}}}{{{\text{Volume}}}} = \frac{{1216}}{{1000}} = 1.216\,g/ml = 1.22g/ml \cr} $$

\[\underset{\begin{smallmatrix} 2\times 122.5g \\ \,\,\,\,\,\,245 \end{smallmatrix}}{\mathop{2KCl{{O}_{3}}}}\,\xrightarrow{\text{heat}}2KCl+\underset{\begin{smallmatrix} 3\times 32g \\ \,\,\,\,96 \end{smallmatrix}}{\mathop{3{{O}_{2}}}}\,\]
$$48$$ $$g$$ of oxygen will be produced from $$122.5\,g$$  of $$KCl{O_3}$$
∴ Amount of $$80\% \,KCl{O_3}$$   needed
$$\eqalign{ & = \frac{{100}}{{80}} \times 122.5 \cr & = 153.12g \cr} $$

$${Z_2}{O_3} + 3{H_2} \to 2Z + 3{H_2}O$$
$$\because \,\,6 \times {10^{ - 3}}g\,{H_2}$$     reduces $$ = 0.{\text{1596}}\,{\text{g}}$$   of $${Z_2}{O_3}$$
$$\eqalign{ & \therefore \,\,1\,g\,{H_2}\,{\text{reduces}} = \frac{{0.1596}}{{6 \times {{10}^{ - 3}}}}g\,{Z_z}{O_3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 26.6\,g\,{\text{of}}\,{Z_2}{O_3} \cr} $$
$$\therefore $$  Equivalent weight of $${Z_2}{O_3} = 26.6$$
Equivalent weight of $$Z$$ + Equivalent weight of $$O$$  $$= 26.6$$
Equivalent weight of $$Z + 8 = 26.6$$
Equivalent weight of $$Z = (26.6 - 8) = 18.6$$
Valency of metal in $${Z_2}{O_3}$$  $$= 3$$
Equivalent weight $$ = \frac{{{\text{Atomic}}\,{\text{weight}}}}{{{\text{Valency}}}}$$
Atomic weight $$ = 18.6 \times 3 = 55.8$$

$${C_3}{H_8} + {O_2} \to C{O_2} + {H_2}O$$
Since the reaction is not balanced hence, mass of reactants and products are different. It is against the law of conservation of mass.

$$\eqalign{ & {\text{Weight of}}{\,^{200}}X = 0.90 \times 200 = 180.00u \cr & {\text{Weight of}}{\,^{199}}X = 0.08 \times 199 = 15.92u \cr & {\text{Weight of}}{\,^{202}}X = 0.02 \times 202 = 4.04u \cr & {\text{Total}}\,{\text{weight = }}\,{\text{199}}{\text{.96}} \approx {\text{200u}} \cr} $$

$$\eqalign{ & 4N{H_3}\left( g \right) + 5{O_2}\left( g \right) \to 4NO\left( g \right) + 6{H_2}O\left( l \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,moles\,\,\,\,\,\,5\,moles\,\,\,\,\,\,4\,moles\,\,\,\,\,6\,moles \cr & {\text{Given}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{1}}\,Mole\,\,\,\,\,\,\,\,1\,Mole \cr & {\text{Reacting}}\,\,\,\,\,\,\,\,0.8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1.2 \cr & {\text{All }}{O_2}\,{\text{consumed being limiting reagent}}{\text{.}} \cr} $$

$$\eqalign{ & {M_1}{V_1} = {M_2}{V_2} \cr & 0.5 \times 100 = 0.1 \times {V_2} \Rightarrow {V_2} = 500\,c{m^3} \cr} $$
Volume of water to be added to $$100\,c{m^3}$$   of solution $$ = 500 - 100 = 400\,c{m^3}$$

$${\text{Molecular weight of}}$$
$$ZnS{O_4}.7{H_2}O$$    $$ = 65 + 32 + \left( {4 \times 16} \right) + 7\left( {18} \right)$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 287 \cr & \therefore \,\,{\text{Percentage weight of }} \cr & Zn = \frac{{65}}{{287}} \times 100 \cr & \,\,\,\,\,\,\,\,\, = 22.65\% \cr} $$

$$\eqalign{ & {\text{Weight of Iron in 67200}} \cr & = \frac{{0.33}}{{100}} \times 67200 \cr & = 221.76 \cr & {\text{Number of atoms of Iron}} \cr & = \frac{{221.76}}{{56}} \cr & = 3.96 \equiv 4 \cr} $$