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

According to law of multiple proportions "if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers".

Specific volume ( volume of $$1g$$  ) cylindrical virus particle
$$\eqalign{ & = 6.02 \times {10^{ - 2}}cc/g \cr & {\text{Radius of virus}}\,\left( r \right) = 7 \times {10^{ - 8}}cm \cr & {\text{Length of virus }}\left( l \right) = 10 \times {10^{ - 8}}cm \cr & {\text{Volume of virus}} \cr & = \pi {r^2}l = \frac{{22}}{7} \times {\left( {7 \times {{10}^{ - 8}}} \right)^2} \times 10 \times {10^{ - 8}} \cr & = 154 \times {10^{ - 23}}cc \cr & {\text{Weight of one virus particle}} \cr & = \frac{{{\text{Volume}}}}{{{\text{Specific volume}}}} \cr & = \frac{{154 \times {{10}^{ - 23}}}}{{6.02 \times {{10}^{ - 2}}}} \cr} $$
∴ Molecular weight of virus = weight of $${N_A}$$  particles
$$\eqalign{ & {\text{ = }}\frac{{154 \times {{10}^{ - 23}}}}{{6.02 \times {{10}^{ - 2}}}} \times 6.023 \times {10^{23}} \cr & = 15400\,g/mol \cr & = 15.4\,kg/mol \cr} $$

$$\eqalign{ & {N_2} + {O_2} \to 2NO \cr & a\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a \cr & CO + \frac{1}{2}{O_2} \to C{O_2} \cr & \left( {10 - a} \right)\,\,\,\frac{{\left( {10 - a} \right)}}{2}\,\,\,\left( {10 - a} \right) \cr & a + \frac{{\left( {10 - a} \right)}}{2} = 7 \cr & \therefore \,\,a = 4 \cr & {\text{volume of}}\,\,CO = 6mL \cr & {\text{Mole}}\,\,\% \,\,{\text{of}}\,\,CO = \frac{{6 \times 100}}{{10}} = 60 \cr} $$

Key Concept In the given problem we have provided practical yield of $$MgO.$$  For calculation of percentage yield of $$MgO,$$  we need theoretical yield of $$MgO.$$  For this we shall use mole concept.
$$\eqalign{ & MgC{O_3}\left( s \right) \to MgO\left( s \right) + C{O_2}\left( g \right)\,\,...{\text{(i)}} \cr & {\text{Moles of}}\,MgC{O_3} \cr & = \frac{{{\text{Weight in gram}}}}{{{\text{Molecular weight}}}} \cr & = \frac{{20}}{{84}} \cr & = 0.238\,mol \cr & {\text{From}}\,Eq.\,{\text{(i)}} \cr & 1\,mole\,{\text{of}}\,MgC{O_3}\,{\text{gives}} = 1\,mol\,MgO \cr & \therefore 0.238\,mole\,MgC{O_3}\,{\text{will}}\,{\text{give}} \cr & = 0.238\,mol\,MgO \cr & = 0.238 \times 40\,g \cr & = 9.52\,g\,MgO \cr & {\text{Now, practical yield of}}\,MgO = 8g \cr & \therefore \,\,\,\% \,{\text{purity}} = \frac{8}{{9.52}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 84\% \cr & {\text{Alternate Method}} \cr & \mathop {MgC{O_3}}\limits_{84\,g} \to \mathop {MgO}\limits_{40\,g} + C{O_2} \cr & \therefore \,\,8\,g\,\,MgO\,{\text{will be form from}}\,\frac{{84}}{5}g \cr & \therefore \,\,\% \,{\text{purity}} = \frac{{84}}{5} \times \frac{{100}}{{20}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 84\% \cr} $$

$$\eqalign{ & {\text{Moles of}}\,{H_2}S{O_4}\,{\text{in}}\,98\,mg\,{\text{of}}\,{H_2}S{O_4} \cr & = \frac{1}{{98}} \times 0.098 \cr & = 0.001 \cr & {\text{Moles of}}\,{H_2}S{O_4}\,{\text{removes}} \cr & = \frac{{3.01 \times {{10}^{20}}}}{{6.02 \times {{10}^{23}}}} \cr & = 0.5 \times {10^{ - 3}} \cr & = 0.0005 \cr & {\text{Moles of}}\,{H_2}S{O_4}\,{\text{left}} \cr & = 0.001 - 0.0005 \cr & = 0.5 \times {10^{ - 3}} \cr} $$

TIPS/Formulae :
Write the reaction for chemical change during reaction and equate moles of products formed.
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br$$     has ionisable $$B{r^ - }$$  ions & $$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$     has ionisable $$S{O_4}^{ - \, - }$$  ion.
Given mixture $$X = 0.02\,{\text{mol}}$$   of $$\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]\,Br$$     and $${\text{0}}{\text{.02}}\,{\text{mol}}$$  of $$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$
Volume $${\text{ = 2}}L$$
∴ Mixture $$X$$ has $${\text{0}}{\text{.02}}\,{\text{mol}}{\text{.}}$$  of $$\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br$$     and $${\text{0}}{\text{.02}}\,{\text{mol}}$$   of $$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$    in $${\text{2}}\,L$$  of solution
∴ Conc. of $$\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]\,Br$$     and $$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$    $$ = 0.01\,mol/L$$    for each of them.
(i) $${\text{1}}\,L$$  ixture of $$X+$$  excess $$AgN{O_3} \to Y$$
$$\mathop {\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br}\limits_{0.01\,mol/L\,\,{\text{soluble}}} + \mathop {AgN{O_3}}\limits_{{\text{excess}}} \to $$       \[\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}S{{O}_{4}} \right]N{{O}_{3}}+\underset{\begin{smallmatrix} \,\,\,\,\,\,\left( Y \right) \\ 0.01\,mol \end{smallmatrix}}{\mathop{AgBr}}\,\]
$$\left[ {A{g^ + } + B{r^ - } \to AgBr} \right]$$
∴ No. of moles of $$Y = 0.01$$
(ii) Also $${\text{1}}\,L$$  mixture of $$X+$$  excess $$BaC{l_2} \to \,Z$$
$$\mathop {\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}}\limits_{0.01\,mol/L\,\,{\text{soluble}}} + \mathop {BaC{l_2}}\limits_{{\text{excess}}} \to $$       \[\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Br \right]C{{l}_{2}}+\underset{\begin{smallmatrix} \,\,\,\,\,\,\left( Z \right) \\ 0.01\,mol \end{smallmatrix}}{\mathop{BaS{{O}_{4}}}}\,\]
$$\left[ {B{a^{ + + }} + \,SO_4^ - \to BaS{O_4}} \right]$$
∴ moles of $$Z = 0.01.$$

$$\eqalign{ & {\text{Molarity}} \cr & = \frac{{Wt \times 1000}}{{MW \times V}} \cr & = \frac{{2.65 \times 1000}}{{106 \times 250}} \cr & = 0.1\,M \cr & {M_1}{V_1} = {M_2}{V_2} \cr & \therefore \,\,10 \times 0.1 = 1000 \times {M_2} = 0.001\,M \cr} $$

$$AgN{O_3} + NaCl \to AgCl + NaN{O_3}$$
No. of moles of $$AgN{O_3} = \frac{{3.4}}{{170}} = 0.02$$
No. of moles of $$NaCl = \frac{{5.85}}{{58.5}} = 0.1$$
Limiting reagent $$ = AgN{O_3}$$
$$1\,mole$$  of $$AgN{O_3}$$  produces $$1\,mole$$  of $$AgCl$$
$$0.02\,mole$$   of $$AgN{O_3}$$  will produce $$0.02\,mole$$   of $$AgCl$$
Weight of $$AgCl$$  produced $$ = 0.02 \times 143.5 = 2.87\,g$$

$$\eqalign{ & {\text{Molarity}}\left( M \right) \cr & = \frac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Volume of solution (in }}L)}} \cr & {\text{or, Molarity}} \cr & = \frac{{{W_B} \times 1000}}{{{M_B} \times {\text{Volume of solution (in }}mL{\text{)}}}} \cr & = \frac{{5.85 \times 1000}}{{58.5 \times 500}}\,\,\left( {\because \,\,{\text{Molar mass of}}\,\,NaCl = 58.5\,g\,mo{l^{ - 1}}} \right) \cr & = 0.2\,mol\,{L^{ - 1}} \cr} $$

74.75 % of chlorine means 74.75 $$g$$  chlorine is present in $$100g$$  of metal chloride.
$$\eqalign{ & {\text{Weight of metal}} \cr & = 100g - 74.75g \cr & = 25.25g \cr & {\text{Equivalent weight}} \cr & = \frac{{{\text{weight of metal}}}}{{{\text{weight of chlorine}}}} \times 35.5 \cr & = \frac{{25.25}}{{74.75}} \times 35.5 \cr & = 12 \cr & {\text{Valency of metal}} \cr & = \frac{{2 \times V.D.}}{{{\text{Equivalent }}wt{\text{. of metal}} + 35.5}} \cr & = \frac{{2 \times 94.8}}{{12 + 35.5}} \cr & = 4 \cr & \therefore {\text{Formula of compound}} = MC{l_4} \cr} $$