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

\[\underset{\begin{smallmatrix} 2\times 108 \\ =216\,g \end{smallmatrix}}{\mathop{2Ag}}\,+S\to \underset{248\,g}{\mathop{A{{g}_{2}}S}}\,\]
$$216{\text{ }}g{\text{ of }}Ag{\text{ forms }}248{\text{ }}g{\text{ of}}\,A{g_2}S$$
$$1.5\,g\,\,{\text{of}}\,\,Ag\,\,{\text{forms}}\,\,\frac{{248}}{{216}} \times 1.5$$       $$ = 1.722\,g\,\,{\text{of}}\,\,A{g_2}S$$
$$\% \,\,{\text{yield of}}\,\,A{g_2}S$$     $$ = \frac{{0.124}}{{1.722}} \times 100 = 7.2\% $$

The change involved is $$Mn{O_4}^ - + {e^ - } \to MnO_4^{2 - }$$
i.e. it involves only one electron
$$Mn{O_4}^ - + {e^ - } \to MnO_4^{2 - }$$
$${\text{Eq}}{\text{.wt}} = \frac{{{\text{Mol}}{\text{.wt}}}}{{{\text{No}}{\text{.}}\,{\text{of}}\,{{\text{e}}^ - }\,{\text{involved}}}} = $$       $$\frac{M}{1} = M\,\left[ {\because {\text{Mol}}{\text{.wt}}{\text{.}} = M} \right]$$

$$\eqalign{ & S{O_2}C{l_2} + 2{H_2}O \to {H_2}S{O_4} + 2HCl \cr & {H_2}S{O_4} + Ba{\left( {OH} \right)_2} \to BaS{O_4} + 2{H_2}O \cr & 2HCl + Ba{\left( {OH} \right)_2} \to BaC{l_2} + 2{H_2}O \cr & {\text{Total}}\,\,moles\,\,{\text{of}}\,Ba{\left( {OH} \right)_2}\,{\text{required}} = 2 \cr} $$

One molecule of oxalic acid dihydrate $${\left( {COOH} \right)_2} \cdot 2{H_2}O$$     contains 6 oxygen atoms.
Number of oxygen atoms in $$1\,mole$$
$$\eqalign{ & = 6 \times 6.023 \times {10^{23}} \cr & = 36.13 \times {10^{23}}\,{\text{atoms}} \cr} $$

Since empirical formula is multiplied by $$n$$  to get molecular formula, $${\left( {C{H_2}{O_2}} \right)_n}$$   where $$n = 1,2,3,...$$   etc.
Hence, among the given options $$C{H_2}{O_2}$$  will give only $${C_2}{H_4}{O_4}$$  as its molecular formula.

An, $$Al{F_3}$$  the number of $$F$$  is 3 for one $$Al{F_3}$$  molecule $$3{F^ - } \equiv 1$$   formula unit of $$Al{F_3}$$
$$3.0 \times {10^{24}}{F^ - } \equiv \frac{1}{3} \times 3.0 \times {10^{24}}Al{F_3}\,units$$

Let atomic masses of $$X$$ and $$Y$$ be $${A_X}$$ and $${A_Y},$$ respectively
$$\eqalign{ & {\text{For}}\,X{Y_2},\,{n_{X{Y_2}}} = 0.1 = \frac{{10}}{{{A_X} + 2{A_Y}}} \cr & {\text{or}}\,\,{A_X} + 2{A_Y} = 100\,\,\,\,\,...{\text{(i)}} \cr & {\text{For}}\,{X_3}{Y_2},\,{n_{{X_3}{Y_2}}} = 0.05 = \frac{9}{{3{A_X} + 2{A_Y}}} \cr & {\text{or}}\,\,3{A_X} + 2{A_Y} = 180\,\,\,\,...{\text{(ii)}} \cr & {\text{On solving Eqs}}{\text{. (i) and (ii), we get,}} \cr & \,\,\,\,\,\,\,\,{A_X} = 40\,g\,mo{l^{ - 1}} \cr & \Rightarrow {A_Y} = 30\,g\,mo{l^{ - 1}} \cr} $$

Formula mass of $$NaCl=$$   Atomic mass of $$Na$$  + Atomic mass of $$Cl$$  $$ = 23.0\,u + 35.5\,u = 58.5\,u$$

In 7 $$g$$  nitrogen, number of molecules
$$ = \frac{{7.0}}{{28}}mol$$
$$ = 0.25 \times {N_A}\,{\text{molecules}}$$
where, $${N_A} = $$  Avogadro number $$ = 6.023 \times {10^{23}}$$
In 2 $$g$$  of $${H_2} = \frac{{2.0}}{2}mol = 1 \times {N_A}\,{\text{molecules}}$$
In 16 $$g$$  of $$N{O_2} = \frac{{16.0}}{{46}}mol$$     $$ = 0.348 \times {N_A}\,{\text{molecules}}$$
In 16 $$g$$  of $${O_2} = \frac{{16}}{{32}}mol$$    $$ = 0.5 \times {N_A}\,{\text{molecules}}$$
Hence, maximum number of molecules are present in 2 $$g$$  of $${H_2}.$$

TIPS/Formulae :
Equivalents of
$$\eqalign{ & {H_2}{C_2}{O_4}.2{H_2}O = {\text{Equivalents}}\,\,{\text{of}}\,\,NaOH \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{At}}\,{\text{equivalence}}\,{\text{point}}} \right) \cr & {\text{Strength}}\,\,{\text{of}}\,\,{H_2}{C_2}{O_4}.2{H_2}O\left( {{\text{in}}\,g/L} \right){\text{ = }}\frac{{6.3}}{{\frac{{250}}{{1000}}}} \cr & {\text{ = 25}}{\text{.2}}\,g/L\, \cr & {\text{Normality}}\,{\text{of}}\,\,{H_2}{C_2}{O_4}.2{H_2}O = \frac{{{\text{strength}}}}{{{\text{Eq}}{\text{.}}\,{\text{wt}}}} \cr & = \frac{{25.2}}{{63}} = 0.4N \cr & \left\{ {{\text{Eq}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of}}\,{\text{oxalic}}\,{\text{acid}} = \frac{{{\text{Mol}}{\text{.}}\,{\text{wt}}}}{2} = \frac{{126}}{2} = 63} \right\} \cr & {\text{Using}}\,{\text{normality}}\,{\text{equation}}: \cr & {N_1}{V_1} = {N_2}{V_2} \cr & \left( {{H_2}{C_2}{O_4}.2{H_2}O} \right)\,\,\,\left( {NaOH} \right) \cr & 0.4 \times 10 = 0.1 \times {V_2}\,or\,{V_2} = \frac{{0.4 \times 10}}{{0.1}} = 40ml. \cr} $$