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

$$\eqalign{ & {\text{Moles}}\,\,{\text{of}}\,{\text{urea}}\,{\text{present}}\,{\text{in}}\,{\text{100}}\,ml\,{\text{of}}\,sol. = \frac{{6.02 \times {{10}^{20}}}}{{6.02 \times {{10}^{23}}}} \cr & \therefore \,M = \frac{{6.02 \times {{10}^{20}} \times 1000}}{{6.02 \times {{10}^{23}} \times 100}} = 0.01M \cr & \left[ {\because \,M\,{\text{ = Moles}}\,\,{\text{of}}\,{\text{solute}}\,{\text{present}}\,{\text{in}}\,{\text{1}}L\,{\text{of}}\,{\text{solution}}} \right] \cr} $$

Among all the given options molarity is correct because the term molarity involve volume which increases on increasing temperature.

Mass of products = Mass of reactants. Hence, mass will remain the same.

$$\mathop {{C_2}{H_2}}\limits_{1\,\,vol.} + \mathop {\frac{5}{2}{O_2}}\limits_{\frac{5}{2}\,\,vol.} \to \mathop {2C{O_2}}\limits_{2\,vol.} + {H_2}O$$
$$1\,vol.\,\,{\text{of}}\,\,{C_2}{H_2}\,\,{\text{requires}} = \frac{5}{2}vol.\,\,{\text{of}}\,\,{O_2}$$
$$2\,vol.\,\,{\text{of}}\,\,{C_2}{H_2}\,\,{\text{requires}}$$     $$ = \frac{5}{2} \times 2 = 5\,\,vol.\,\,{\text{of}}\,\,{O_2}$$

$$95\% \,{H_2}S{O_4}$$   by weight means $$100g\,{H_2}S{O_4}$$   solution contains $$95g\,{H_2}S{O_4}$$   by mass.
Molar mass of $${H_2}S{O_4} = 98g\,mo{l^{ - 1}}$$
Moles in $$95g = \frac{{95}}{{98}} = 0.969\,mole$$
$$\eqalign{ & {\text{Volume of 100}}g\,\,{H_2}S{O_4} \cr & = \frac{{{\text{mass}}}}{{{\text{density}}}} = \frac{{100g}}{{1.834g\,c{m^{ - 3}}}} \cr & = 54.52\,c{m^3} \cr & = 54.52 \times {10^{ - 3}}L \cr & {\text{Molarity}} = \frac{{{\text{Moles of solute}}}}{{{\text{Volme of solute in}}\,L}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.969}}{{54.52 \times {{10}^{ - 3}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 17.8\,M \cr} $$

$$\eqalign{ & m{\text{ - }}eq.\,FeS{O_4}{\left( {N{H_4}} \right)_2}S{O_4}.6{H_2}O \cr & = m{\text{ - }}eq.\,{\text{of}}\,KMn{O_4} \cr & \left( {n = 1} \right) \cr & \frac{W}{{392}} \times 1 \times 1000 \cr & = 0.1 \times 50 \cr & {\text{Hence, }}\% {\text{ purity of Mohr's salt}} \cr & {\text{ = }}\frac{{1.96}}{{2.5}} \times 100 = 78.4\% \,\,\,\,\,\,\,\,\,W = 1.96\,g \cr} $$

$$\eqalign{ & Fe\left( {{\text{no}}{\text{.}}\,{\text{of}}\,{\text{moles}}} \right) = \frac{{558.5}}{{55.85}} = 10\,{\text{moles}} \cr & C\left( {{\text{no}}{\text{.}}\,{\text{of}}\,{\text{moles}}} \right)\,{\text{in}}\,{\text{60}}\,{\text{g}}\,{\text{of}}\,C = \frac{{60}}{{12}} = 5\,{\text{moles}}{\text{.}} \cr} $$

On balancing the given equations we get
$$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to $$       $$2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$$
$${\text{So}},\,\,x = 2,\,y = 5\,\,\& \,z = 16$$

$$\eqalign{ & CO + \frac{1}{2}{O_2} \to C{O_2}; \cr & C{O_2} + 2KOH \to {K_2}C{O_3} + {H_2}O \cr & {\text{Moles of}}\,KOH = \frac{{28}}{{56}} = 0.50 \cr & {\text{It corresponds to}}\,0.25\,mol\,{\text{of}}\,C{O_2} \cr} $$
$${\text{Hence}}\,mol\,{\text{of}}\,CO = 1 - 0.25$$       $$ = 0.75 \equiv {\text{mole of}}\,C{O_2}\,{\text{formed}}$$
$$\eqalign{ & Mol\,{\text{of}}\,KOH\,{\text{requred}} = 2 \times 0.75 = 1.5 \cr & = 1.5 \times 56 = 84\,g \cr} $$

$$\eqalign{ & {\text{No}}{\text{. of atoms in}}\,\,1\,mole\,\,{\text{of}}\,\,PC{l_3} = 4 \cr & {\text{No}}{\text{. of atoms in }}1.4{\text{ }}moles{\text{ of}}\,\,PC{l_3} \cr & = 4 \times 1.4 \times 6.023 \times {10^{23}} \cr & = 3.372 \times {10^{24}} \cr} $$