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

$${\text{No}}{\text{.}}\,\,{\text{of}}\,\,{e^ - }\,\,{\text{in}}\,\,C = 6\,\,{\text{and}}\,\,{\text{in}}\,\,O = 8$$
$$\therefore \,\,{\text{Total}}\,\,{\text{no}}{\text{.}}\,\,{\text{of}}\,\,{e^ - }\,\,{\text{in}}\,\,C{O_2} = $$       $$6\, + 8 \times 2 = 22$$

$$\eqalign{ & {\text{Molar mass of}} \cr & CuS{O_4} = 63.5 + 32 + 64 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 159.5 \cr & {\text{Moles of}}\,\,CuS{O_4} = \frac{{80}}{{159.5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.50 \cr & {\text{Volume of solution}} = 3\,L \cr & {\text{Molarity}} = \frac{{{\text{Moles of solute}}}}{{{\text{VoIume of solution}}\,\,{\text{in}}\,\,L}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.50}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.167\,mol\,{L^{ - 1}} \cr} $$

Mass of compound taken $$ = 0.48\,g$$
Mass of boron in sample $$ = 0.192\,g.$$
Mass of oxygen in sample $$ = 0.288\,g$$
Percentage of $$B = \frac{{0.192}}{{0.48}} \times 100 = 40\% $$
Percentage of $$O = \frac{{0.288}}{{0.48}} \times 100 = 60\% $$

$$\eqalign{ & {\text{Relative atomic mass}} \cr & {\text{ = }}\frac{{{\text{Mass of one atom of the element}}}}{{{{\frac{1}{{12}}}^{th}}\,{\text{part of the mass of one atom of Carbon}}\, - 12}} \cr & {\text{or}}\,\,\frac{{{\text{Mass of one atom of the element}}}}{{{\text{mass of one atom of the}}\,C - 12}} \times 12 \cr & {\text{Now if we use}}\,\,\frac{1}{6}\,\,{\text{in place of}}\,\,\frac{1}{{12}}\,\,{\text{the formula becomes}} \cr & {\text{Relative atomic mass}} = \frac{{{\text{Mass of one atom of element}}}}{{{\text{Mass of one atom of carbon}}}} \times 6 \cr & \therefore \,{\text{Relative atomic mass decrease twice}}{\text{.}} \cr} $$

$$2Al\left( s \right) + 6HCl\left( {aq} \right) \to $$     $$2A{l^{3 + }}\left( {aq} \right) + 6C{l^ - }\left( {aq} \right) + 3{H_2}\left( g \right)$$
$$\because \,6\,{\text{moles}}\,{\text{of}}\,HCl\,{\text{produces}}\, = $$       $$3\,{\text{moles}}\,{\text{of}}\,{H_2}$$
$$ = 3 \times 22.4\,L\,{\text{of}}\,{H_2}\,{\text{at}}\,S.T.P$$
$$\therefore 1\,{\text{mole}}\,{\text{of}}\,HCl\,{\text{produces}} = $$       $$\frac{{3 \times 22.4}}{6}L\,\,{\text{of}}\,{H_2}\,{\text{at}}\,{\text{S}}{\text{.T}}{\text{.P}}$$
$$ = 11.2\,L\,\,{\text{of}}\,\,{H_2}\,{\text{at}}\,\,{\text{STP}}$$

Suppose the molecular weight of enzyme $$ = x$$
$$0.5\% $$  by weight means in $$100g$$  of enzyme weight of $$Se = 0.5g$$
$$\therefore $$  In $$xg$$  of enzyme weight of $${\text{ }}Se = \frac{{0.5}}{{100}} \times x$$
$$\eqalign{ & {\text{Hence,}}\,{\text{78}}{\text{.4 = }}\frac{{0.5 \times x}}{{100}} \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 15680 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.568 \times {10^4} \cr} $$

For equivalent weight of $$MnS{O_4}$$   to be half of its molecular weight, change in oxidation state must be equal to 2. It is possible only when oxidation state of $$Mn$$  in product is $$+ 4.$$  Since oxidation state of $$Mn$$  in
$$\eqalign{ & MnS{O_4}\,{\text{is}}\,{\text{ + }}\,{\text{2}}{\text{.}}\,\,{\text{So,}}\,Mn{O_2}\,{\text{is}}\,{\text{correct}}\,{\text{answer}}{\text{.}} \cr & {\text{In}}\,Mn{O_2},O.S.\,{\text{of}}\,Mn\, = + 4 \cr & \therefore {\text{Change}}\,{\text{in}}\,O.S.\,{\text{of}}\,Mn = + 4 - \left( { + 2} \right) = + 2 \cr} $$

No. of moles of $$34\,g$$  of $$N{H_3} = \frac{{34}}{{17}} = 2$$
No. of molecules $$ = 2 \times 6.023 \times {10^{23}}$$
No. of atoms in one molecule of $$N{H_3} = 4$$
No. of atoms in $$2 \times 6.023 \times {10^{23}}$$    molecules of $$N{H_3}$$
$$\eqalign{ & = 4 \times 2 \times 6.023 \times {10^{23}} \cr & = 48.18 \times {10^{23}} \cr} $$

$${\text{One molecule of sugar}}$$     $$\left( {{C_{12}}{H_{22}}{O_{11}}} \right) = 45\,{\text{atoms}}$$
$$\eqalign{ & {\text{Number of moles of sugar}} \cr & = \frac{{1000}}{{342}} \cr & = 2.92 \cr & {\text{Number of molecules}} \cr & = 2.92 \times 6.023 \times {10^{23}} \cr & = 17.60 \times {10^{23}}\,{\text{molecules}} \cr & {\text{Number of atoms}} \cr & = 45 \times 17.60 \times {10^{23}} \cr & = 7.92 \times {10^{25}}\,{\text{atoms}} \cr} $$

$$\eqalign{ & C\,\,\,\,\,\,54.55\,\,\,\,\,\,\frac{{54.55}}{{12}} = 4.5\,\,\,\,\,\,\,\frac{{4.5}}{{2.27}} = 2 \cr & H\,\,\,\,\,9.09\,\,\,\,\,\,\,\,\frac{{9.09}}{1} = 9.09\,\,\,\,\,\,\,\,\frac{{9.09}}{{2.27}} = 4 \cr & O\,\,\,\,\,36.36\,\,\,\,\,\frac{{36.36}}{{16}} = 2.27\,\,\,\,\,\frac{{2.27}}{{2.27}} = 1 \cr} $$
Hence empirical formula of the compound $$ = {C_2}{H_4}O$$