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

\[\underset{232\,g}{\mathop{F{{e}_{3}}{{O}_{4}}}}\,+4CO\to \underset{\begin{smallmatrix} 56\times 3 \\ =168\,g \end{smallmatrix}}{\mathop{3Fe}}\,+4C{{O}_{2}}\]
$$3\,moles$$   of $$Fe$$  is produced from $$1\,mole$$  of $$F{e_3}{O_4}$$
$$168\,g$$  of $$Fe$$  is produced from $$232\,g$$  of $$F{e_3}{O_4}$$
$$3\,kg$$  of $$Fe$$  will be produced from $$\frac{{232}}{{168}} \times 3000\,g$$    $$ = 4142.8\,g$$   or $$4.14\,kg$$   of $$F{e_3}{O_4}$$

If Avogadro number $${N_A},$$  is changed from $$6.022 \times {10^{23}}mo{l^{ - 1}}$$    to $$6.022 \times {10^{20}}mo{l^{ - 1}},$$    this would change the mass of one mole of carbon.
$$\because \,\,1\,mole$$   of carbon has mass = 12 $$g$$
or $$6.022 \times {10^{23}}$$   atoms of carbon have mass = 12 $$g$$
∴ $$6.022 \times {10^{20}}$$   atoms of carbon have mass
$$\eqalign{ & = \frac{{12}}{{6.022 \times {{10}^{23}}}} \times 6.022 \times {10^{20}} \cr & = 0.012g \cr} $$

$$\eqalign{ & 1\,mole\,{\text{of}}\,C{O_2} \cr & = 44\,g\,{\text{of}}\,C{O_2} \cr & = 6.023 \times {10^{23}}molecules \cr & \therefore \,\,4.4\,g\,{\text{of}}\,C{O_2} \cr & = 0.1\,mole\,{\text{of}}\,C{O_2} \cr & = 6.023 \times 0.1 \times {10^{23}}molecules \cr & = 6.023 \times {10^{22}}molecules\, \cr & = 6.023 \times {10^{22}}molecules\,{\text{of}}\,{O_2} \cr & = 2 \times 6.023 \times {10^{22}}atoms\,{\text{of}}\,O \cr & = 1.2 \times {10^{23}}atoms\,{\text{of}}\,O \cr} $$

$$\eqalign{ & {\text{Given}} \cr & {\text{mass of solute}}\left( w \right) = 120\,g \cr & {\text{mass of solvent}}\left( w \right) = 1000\,g \cr & Mol.{\text{ mass of solute}} = 60\,g \cr & {\text{density of solution}} = 1.12\,g/mL \cr & {\text{From the given data,}} \cr & {\text{Mass of solution}} \cr & = 1000 + 120 = 1120\,g \cr & \because \,\,d = \frac{{Mol.\,mass}}{V}\,{\text{or}}\,\,V = \frac{{Mol.\,mass}}{d} \cr & {\text{Volume of solution}} \cr & V = \frac{{1120}}{{1.12}} = 1000\,mL \cr & {\text{or = 1 litre}} \cr & {\text{Now molarity}} \cr & \left( M \right) = \frac{W}{{Mol.\,mass \times V\left( {lit} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{120}}{{60 \times 1}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 2M \cr} $$

Percentage of element $$'X' = 50,$$
moIe ratio $$ = \frac{{50}}{{20}} = 2.5$$
Percentage of element $$'Y' = 50,$$   mole ratio $$ = \frac{{50}}{{40}} = 1.25$$
For $$'X,'$$  simple ratio $$ = \frac{{2.5}}{{1.25}} = 2$$
For $$'Y,'$$  simple ratio $$ = \frac{{1.25}}{{1.25}} = 1$$
Formula $$ = {X_2}Y$$

According to Gay Lussac's law of gaseous volume, when gases combine or are produced in a chemical reaction, they do so in simple ratio by volume provided all gases are at same temperature and pressure.
$$\mathop {2{H_2}}\limits_{2\,vol.} + \mathop {{O_2}}\limits_{1\,vol.} \to \mathop {2{H_2}O}\limits_{2\,vol.} $$

$${\text{Molar mass of}}\,\,A{l_2}{O_3}$$     $$ = 2 \times 27 + 3 \times 16 = 102\,g$$
$$0.051\,g\,\,{\text{of}}\,\,A{l_2}{O_3}$$    $$ = \frac{{0.051}}{{102}} = 0.0005\,mol$$
$$1\,mol\,\,{\text{of}}\,\,A{l_2}{O_3}\,\,{\text{contains}}$$      $$2 \times 6.023 \times {10^{23}}\,\,A{l^{3 + }}\,ions$$
$$0.0005\,mol\,\,{\text{of}}\,\,A{l_2}{O_3}\,\,{\text{contains}}$$      $$2 \times 0.0005 \times 6.023 \times {10^{23}}$$
$$A{l^{3 + }}\,ions = 6.023 \times {10^{20}}\,A{l^{3 + }}\,ions$$

$$\eqalign{ & {X_2}{O_3} \Rightarrow {X^{3 + }}{O^{2 - }} \cr & {X_2}C{l_3} \Rightarrow {X^{3 + }}C{l^{2 - }} \cr & {X_2}{\left( {S{O_4}} \right)_3} \Rightarrow {X^{3 + }}S{O_4}^{2 - } \cr & XP{O_4} \Rightarrow {X^{3 + }}P{O_4}^{3 - } \cr} $$
Because $$C{l^{2 - }}$$ does not exist. So, $${X_2}C{l_3}$$  is incorrect. The correct formula should be $$XC{l_3}.$$

$$\eqalign{ & {\text{No}}{\text{. of}}\,\,O\,\,{\text{atoms}}\,\,{\text{in}}\,\,C{O_2} = 2 \cr & {\text{Molar mass of}}\,\,C{O_2} = 44\,g \cr & 44\,\,g \equiv 1\,mol \Rightarrow 22\,g \equiv 0.5\,mol \cr} $$
$$1\,mole\,\,{\text{of}}\,\,C{O_2}\,\,{\text{contains}}$$      $$ = 2 \times 6.023 \times {10^{23}}O\,\,{\text{atoms}}$$
$$0.5\,mole\,\,{\text{of}}\,\,C{O_2}\,\,{\text{contains}}$$      $$= 6.023 \times {10^{23}}O\,\,{\text{atoms}}$$
$$1\,mole\,\,{\text{of}}\,\,CO\,\,{\text{contains}}$$     $$ = 6.023 \times {10^{23}}O\,\,{\text{atoms}}$$
$${\text{Mass of}}\,\,1\,mole\,\,{\text{of}}\,\,CO$$     $$ = 12 + 16 = 28\,g$$

$$\eqalign{ & \mathop {NaCl}\limits_{x\,g} + \mathop {{H_2}S{O_4}}\limits_{9.8\,g} \to \mathop {NaHS{O_4}}\limits_{12\,g} + \mathop {HCl}\limits_{2.75\,g} \cr & x + 9.8 = 12 + 2.75 \Rightarrow x = 4.95 \cr} $$