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

$$\mathop {CaC{O_3}}\limits_{100\,g} + \mathop {2HCl \to }\limits_{2 \times 36.5 = 73\,g} CaC{l_2} + {H_2}O + C{O_2}$$
$$100\,g\,\,{\text{of}}\,\,CaC{O_3}$$    $${\text{requires}}\,\,73\,g\,\,{\text{of}}\,\,HCl$$
$$\therefore 50\,g\,\,{\text{of}}\,\,CaC{O_3}\,\,{\text{requires}}$$      $$\frac{{73}}{{100}} \times 50 = 36.5\,g\,\,{\text{of}}\,\,HCl$$

$$\eqalign{ & ppm = \frac{{wt.\,{\text{of}}\,{\text{solute}}}}{{wt.\,{\text{of solvent}}}} \times {10^6} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.2}}{{500}} \times {10^6} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 400 \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{I_2} + 2C{l_2} \to ICl + IC{l_3} \cr & {\text{No}}{\text{. of moles}}\,\,\,\,\,\frac{{25.4}}{{254}}\,\,\,\frac{{14.2}}{{71}}\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & {\text{initially}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.1\,\,\,\,\,\,\,0.2\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & {\text{No}}{\text{. of moles}}\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0.1\,\,\,\,\,\,\,\,\,0.1 \cr & {\text{after reaction}} \cr} $$

(i) All non-zero digits are significant.
(ii) Non-zero digits to the right of the decimal point are significant.
(iii) Zeroes to the left of the first non-zero digit in a number are not significant.
So, the number of significant figures for the numbers $$161\,cm,\,0.161\,cm,\,$$   and $$0.0161\,cm$$   are same, i.e. 3.

$$\eqalign{ & {\text{For a one mole of the oxide}} \cr & {\text{Moles of M = 0}}{\text{.98, Moles of}}\,{O^{2 - }} = 1 \cr & {\text{Let moles of}}\,{M^{3 + }} = x \cr & {\text{Moles of}}\,{M^{2 + }} = 0.98 - x \cr & {\text{on balancing charge}} \cr & \left( {0.98 - x} \right) \times 2 + 3x - 2 = 0\,\,\, \Rightarrow \,x = 0.04 \cr & \% \,{\text{of}}\,{M^{3 + }}\frac{{0.04}}{{0.98}} \times 100 = 4.08\% \cr} $$

Element	Percentage	Atomic mass	Relative no. of atoms	Simplest ratio
$$C$$	20%	12	1.66	1
$$H$$	6.7%	1	6.7	4
$$N$$	46.7%	14	3.33	2
$$O$$	26.6%	16	1.66	1

Empirical formula = Molecular formula
$$ = C{H_4}{N_2}O\,\,\,{\text{or}}\,\,N{H_2}CON{H_2}$$
$${H_2}NCON{H_2} + {H_2}NCON{H_2}$$       \[\xrightarrow{\Delta }\,\underset{\text{biuret,}}{\mathop{{{H}_{2}}NCONHCON{{H}_{2}}}}\,+N{{H}_{3}}\]
When an aqueous solution of biuret is treated with dilute sodium hydroxide and a drop of copper sulphate, a violet colour is produced. This test is known as biuret test, and is characteristic of compounds having the group \[-CONH-\]

Element	Relative mass	Relative mole	Simplest whole number ratio
$$C$$	6	$$\frac{6}{{12}} = 0.5$$	1
$$H$$	1	$$\frac{1}{1} = 1$$	2

$$\eqalign{ & {\text{So}},\,X = 1,\,Y = 2 \cr & {\text{Equation for combustion of}}\,{{\text{C}}_X}{H_Y} \cr & {C_X}{H_Y} + \left( {X + \frac{Y}{4}} \right){O_2} \to XC{O_2} + \frac{Y}{2}{H_2}O \cr & {\text{Oxygen atoms required}} = 2\left( {X + \frac{Y}{4}} \right) \cr & {\text{As mentioned,}} \cr & 2\left( {X + \frac{Y}{4}} \right) = 2Z \Rightarrow \left( {1 + \frac{2}{4}} \right) = Z \Rightarrow Z = 1.5 \cr & \therefore \,\,{\text{molecule can be written as}} \cr & {C_X}{H_Y}{O_Z} \cr & {C_1}{H_2}{O_{\frac{3}{2}}} \cr & \Rightarrow {C_2}{H_4}{O_3} \cr} $$

At $$S.T.P.$$  22.4 litre of gas contains $$6.023 \times {10^{23}}$$   molecules
∴ molecules in 8.96 litre of gas
$$\eqalign{ & = \frac{{6.023 \times {{10}^{23}} \times 8.96}}{{22.4}} \cr & = 24.08 \times {10^{22}} \cr} $$

\[\underset{\begin{smallmatrix} 207+16 \\ =223 \\ \left( 1\,mol \right) \end{smallmatrix}}{\mathop{PbO}}\,+\underset{\begin{smallmatrix} 2\times 36.5 \\ =73 \\ \left( 2\,mol \right) \end{smallmatrix}}{\mathop{2HCl}}\,\to \underset{\left( 1\,mol \right)}{\mathop{PbC{{l}_{2}}}}\,+{{H}_{2}}O\]
$$\eqalign{ & {\text{Mole}}\,{\text{of}}\,PbO = \frac{{6.5}}{{223}} = 0.029 \cr & {\text{Mole of}}\,HCl = \frac{{3.2}}{{36.5}} = 0.087 \cr} $$
Since, $$1\,mole$$  of $$PbO$$  reacts with $$2\,mole$$  of $$HCl,$$  thus in this reaction $$PbO$$  is the limiting reagent.
Hence, $$1\,mole$$  of $$PbO$$  forms $$ = 1\,mole\,\,{\text{of}}\,\,PbC{l_2}$$
$$0.029\,mole\,$$   of $$PbO$$  will form $$ = 0.029\,\,mole\,\,{\text{of}}\,\,PbC{l_2}$$

$${\text{Number of moles}}$$
$$\frac{{36}}{{18}}{\text{of}}\,\,{H_2}O;\frac{{28}}{{28}}{\text{of}}\,\,CO;$$     $$\frac{{46}}{{46}}\,{\text{of}}\,\,{C_2}{H_5}OH\,\,{\text{and}}\,\,\frac{{54}}{{108}}\,{\text{of}}\,\,{N_2}{O_5}$$
$$\therefore \,\,2\,moles\,\,{\text{of}}\,\,{H_2}O,$$     $$1\,mole\,\,{\text{of}}\,\,CO,1\,mole\,\,{\text{of}}\,\,{C_2}{H_5}OH$$       $${\text{and }}0.5\,\,mole\,\,{\text{of}}\,\,{N_2}{O_5}.$$
$$\therefore {\text{Largest number of molecules is in (A)}}{\text{.}}$$