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

Let $$x$$ $$g$$ of $$L{i_2}C{O_3}$$  and $$\left( {1 - x} \right)\,g$$   of $${M_2}C{O_3}$$  present in given mixture
2 × Total moles of carbonates = moles of $$HCl$$  and
$$\eqalign{ & 2 \times \left( {\frac{x}{{74}} + \frac{{1 - x}}{{2M + 60}}} \right) \cr & = 21.6 \times 0.5 \times {10^{ - 3}} \cr & x = 0.20 \cr & \because \,\,\frac{x}{{74}} = \frac{{1 - x}}{{2M + 60}} \cr & M = 118 \cr} $$

TIPS/Formulae :
Balance the reaction by ion electron method.
Oxidation reaction:  $${C_2}O_4^{ - 2} \to 2C{O_2} + 2\left. {{e^ - }} \right] \times 5$$
Reduction reaction :
$$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}\left. O \right] \times 2$$
Net reaction :
$$2MnO_4^ - + 16{H^ + } + 5{C_2}O_4^{2 - } \to 2M{n^2} + 10C{O_2} + 8{H_2}O$$

Percentage (by mass) of elements given in the body of a healthy human adult is :-
$$\eqalign{ & {\text{Oxygen}} = 61.4\% ,\,{\text{Carbon}} = 22.9\% , \cr & {\text{Hydrogen}} = 10.0\% \,{\text{and}}\,{\text{Nitrogen}} = 2.6\% \cr & \because {\text{Weight}}\,{\text{of}}\,{\text{person}} = 75\,kg \cr & \therefore \,\,{\text{Mass}}\,{\text{due}}\,{\text{to}}\,{}^1H\,{\text{is}} = 75 \times \frac{{10}}{{100}} = 7.5kg \cr & {\text{If}}\,{}^1H\,{\text{atoms}}\,{\text{are}}\,{\text{replaced}}\,{\text{by}}\,{}^2H\,{\text{atoms,}} \cr & {\text{mass}}\,{\text{gain}}\,{\text{by}}\,{\text{person}}\,{\text{would}}\,{\text{be}} = 7.5\,kg \cr} $$

TIPS/Formulae :
(i) Find oxidation state of $$N$$  in $${N_2}{H_4}.$$
(ii) Find change in oxidation number with the help of number of electrons given out during formation of compound $$Y$$ .
$${N_2}{H_4} \to Y + 10{e^ - },$$     Calculation of $$O.S$$  of $$N$$ in $${N_2}{H_4}:$$
$$2x + 4 = 0 \Rightarrow x = - 2$$
The two nitrogen atoms will balance the charge of $$10e$$ . Hence oxidation state of $$N$$ will increase by + 5, i.e. from -2 to +3.

$$BaC{O_3} \to BaO + C{O_2}$$
$$192\,g$$  of $$BaC{O_3}$$  gives $$1$$ $$mole$$  of $$C{O_2} = 22.4\,L$$
$$9.85\,g$$  of $$BaC{O_3}$$  will give $$0.05$$  $$mol$$  of $$C{O_2}$$  which is equal to $$1.12$$  $$litre.$$

Element	Percentage	Molar ratio	Relative ratio	Whole no. ratio
$$Mg$$	21.9	$$\frac{{21.9}}{{24}} = 0.91$$	1	2
$$P$$	27.8	$$\frac{{27.8}}{{31}} = 0.90$$	1	2
$$O$$	50.3	$$\frac{{50.3}}{{16}} = 3.14$$	3.48	7

Formula of the compound $$ = M{g_2}{P_2}{O_7}$$

$${\text{pico}} = {10^{ - 12}},{\text{femto}} = {10^{ - 15}},$$       $${\text{milli}} = {10^{ - 3}},{\text{giga}} = {10^9}$$

TIPS/Formulae:
(i) Write balanced chemical equation for chemical change.
(ii) Find limiting reagent.
(iii) Amount of product formed will be determined by amount of limiting reagent.
The balanced equation is :
\[\underset{\begin{smallmatrix} No.\,of \\ moles: \end{smallmatrix}}{\mathop {}}\,\,\,\underset{\begin{smallmatrix} \,\,\,0.5 \\ 3\,mole \end{smallmatrix}}{\mathop{3BaC{{l}_{2}}}}\,+\underset{\begin{smallmatrix} \,\,\,0.2 \\ 2\,mole \end{smallmatrix}}{\mathop{2N{{a}_{3}}P{{O}_{4}}}}\,\to \underset{\begin{smallmatrix} \\ 1\,mole \end{smallmatrix}}{\mathop{B{{a}_{3}}\left( P{{O}_{4}} \right)}}\,+6NaCl\]
Limiting reagent is $$N{a_3}P{O_4}\left( {0.2\,mol} \right),\,BaC{l_2}$$     is in excess.
From the above equation :
$$\eqalign{ & 2.0\,{\text{moles}}\,\,{\text{of}}\,N{a_3}P{O_4}\,{\text{yields}}\,\,B{a_3}{\left( {P{O_4}} \right)_2} = 1\,{\text{mole}} \cr & \therefore {\text{0}}{\text{.2}}\,{\text{moles of}}\,N{a_3}P{O_4}\,{\text{will}}\,{\text{yields}}\,B{a_3}{\left( {P{O_4}} \right)_2} = \frac{1}{2} \times 0.2 \cr & = 0.1\,{\text{mol}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Let }}mass{\text{ of}}\,A{g_2}O = x\,g \cr & A{g_2}O \to 2Ag + \frac{1}{2}{O_2} \cr & xg \cr & 232\,g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2} \times 32 = 16\,g\,{O_2} \cr & \frac{{16}}{{232}} \times x = 0.104 \Rightarrow x = 1.508\,g \cr & \% \,\,{\text{of}}\,A{g_2}O = \frac{{1.508}}{{1.6}} \times 100 = 94.25\% \cr} $$

$$N{a_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2NaCl$$
$${\text{No}}{\text{. of moles of}}\,\,BaS{O_4}$$     $$ = \frac{w}{M} = \frac{{10}}{{233}} = 0.0429$$
$$\therefore \,\,{\text{No}}{\text{. of moles of}}\,\,N{a_2}S{O_4}\,\,{\text{needed}}$$       $$ = \frac{{M \times V}}{{1000}}$$
$${\text{or}}\,\,\,0.0429 = \frac{{5 \times V}}{{1000}} \Rightarrow V = 8.58\,mL$$