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

$$\eqalign{ & {\text{Average atomic mass of chlorine}} \cr & = \frac{{3 \times 35 + 37 \times 1}}{4} \cr & = 35.5 \cr} $$

$$1\,mole\,\,{\text{of}}\,\,C{H_4} = 16\,g$$
$$16\,g\,\,{\text{of}}\,\,C{H_4}\,\,{\text{contains}}$$     $$10 \times 6.023 \times {10^{23}}\,{\text{electrons}}$$
$$1.6\,g\,\,{\text{of}}\,\,C{H_4}\,\,{\text{contains}}$$     $$\frac{{10}}{{16}} \times 1.6 \times 6.023 \times {10^{23}}$$
$$ = 6.023 \times {10^{23}}\,{\text{electrons}}$$

$$\mathop {{H_2}}\limits_{2\,g} + \mathop {C{l_2}}\limits_{71g} \to \mathop {2HCl}\limits_{73\,g} $$
$$2\,g$$  of $${H_2}$$  reacts with $$71\,g$$  of $$C{l_2}$$
$$100\,g$$  of $${H_2}$$  will react with $$\frac{{71}}{2} \times 100 = 3550\,g$$     of $$C{l_2}$$
Hence, $$C{l_2}$$  is the limiting reagent.
$$71\,g$$  of $$C{l_2}$$  produces $$73\,g$$  of $$HCl$$
$$100\,g$$  of $$C{l_2}$$  will produce $$\frac{{73}}{{71}} \times 100 = 102.8\,g$$     of $$HCl$$

$$Mg{\left( {OH} \right)_2} + 2HCl \to MgC{l_2} + 2{H_2}O$$
No. of moles of $$Mg{\left( {OH} \right)_2}$$   required for $$2\,moles$$   of $$HCl = 1$$
No. of moles of $$Mg{\left( {OH} \right)_2}$$   required for $$1\,mole$$   of $$HCl = 0.5$$
Mass of $$0.5\,mol$$   of $$Mg{\left( {OH} \right)_2} = 58.33 \times 0.5 = 29.16\,g$$

$$\eqalign{ & n = \frac{{{\text{Molecular mass}}}}{{{\text{Empirical formula mass}}}} \cr & \,\,\,\,\, = \frac{{180}}{{30}} \cr} $$
$$\,\,\,\,\,\,\, = 6$$    $$\left( {\because \,\,{\text{Empirical formula mass of}}\,\,C{H_2}O = 30} \right)$$
$$\therefore \,\,{\text{Molecular formula}} = {\left( {C{H_2}O} \right)_6} = {C_6}{H_{12}}{O_6}$$

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [ possessing minimum number of moles and gets completely used up in the reaction ]. The limiting reagent gives the moles of product formed in the reaction.
$${}_{\text{Initial vol}\text{.}}^{}{}_{22.4L}^{H_2\left(g\right)+}{}_{11.2L}^{C{l}_2\left(g\right)\to }{}_{2mol}^{2HCl\left(g\right)}$$
$$\because \,\,22.4\,L\,{\text{ }}$$  volume at $$STP$$  is occupied by
$$C{l_2} = 1\,mole$$
$$\therefore 11.2\,L\,$$  volume will be occupied by
$$\eqalign{ & C{l_2} = \frac{{1 \times 11.2}}{{22.4}}mol \cr & = 0.5\,mol \cr} $$
$$22.4\,L\,{\text{ }}$$  volume at $$STP$$  is occupied by $${H_2} = 1\,mol$$
Thus,   $$\mathop {{H_2}\left( g \right)}\limits_{1\,mol} + \mathop {C{l_2}\left( g \right)}\limits_{0.5\,mol} \to 2HCl\left( g \right)$$
Since, $$C{l_2}$$  possesses minimum number of moles, thus it is the limiting reagent.
As per equation,
$$\eqalign{ & 1\,mole\,{\text{of}}\,C{l_2} = 2\,moles\,{\text{of}}\,HCl \cr & \therefore \,\,0.5\,mole\,{\text{of}}\,C{l_2} = 2 \times 0.5\,mole\,{\text{of}}\,HCl \cr & = 1.0\,mole\,{\text{of}}\,HCl \cr} $$
Hence, $$1.0\,mole$$  of $$HCl\left( g \right)$$  is produced by $$0.5\,mole$$   of $$C{l_2}\left[ {{\text{or}}\,\,11.2\,L} \right].$$

$$\eqalign{ & 2N{H_3} + 5{F_2} \to {N_2}{F_4} + 6HF \cr & 2 \times 17\,g\,N{H_3}\,\,{\text{gives}}\,\,66\,g\,{N_2}{F_4} \cr & 2\,g\,{\text{will give}} - \frac{{66}}{{34}} \times 2 = 3.88\,g\,{N_2}{F_4} \cr & \% \,{\text{yield}} = \frac{{3.56}}{{3.88}} \times 100 = 91.75\% \cr} $$

$$\eqalign{ & {\text{Moles of}}\,C = {n_{C{O_2}}} = \frac{{2.64}}{{44}} = 0.06 \cr & \Rightarrow {\text{mass of}}\,C = 0.72 \cr} $$
$${\text{Mole of}}\,H = 2 \times {\text{Moles of}}\,{H_2}O$$       $$ = 2 \times \frac{{1.26}}{{18}} = 0.14 \Rightarrow {\text{mass of}}\,H = 0.14$$
$$\eqalign{ & {\text{Compound does not contains oxygen}}{\text{.}} \cr & {\text{So}}\,\,EF \to {C_{0.06}}{H_{0.14}} \Rightarrow {C_3}{H_{17}} \cr & \Rightarrow {\text{Lowest}}\,M.M. = 43 \cr} $$

Let the alcohol be $$ROH$$  and $$x$$ its molecular weight
$$\mathop {ROH}\limits_{x\,g} + C{H_3}MgI \to \mathop {C{H_4}}\limits_{16\,g} + ROMgI$$
$$\frac{{4.12}}{{1000}}g$$   of alcohol will produce
$$\frac{{16}}{x} \times \frac{{4.12}}{{1000}}g$$    of methane
Methane actually obtained is $$ = \frac{{16 \times 1.12}}{{22400}}g$$
$$\eqalign{ & \therefore \,\,\frac{{16 \times 4.12}}{{x \times 1000}} \cr & = \frac{{16 \times 1.12}}{{22400}} \cr & = x = 82.4 \cr} $$

In Hall and Heroult process,
$$2\,A{l_2}{O_3} + 4C \to 4\,Al + 2C{O_2} + 2CO$$
but for the removal of only $$C{O_2},$$  following equation is possible.
\[2\,A{{l}_{2}}{{O}_{3}}+\underset{\begin{smallmatrix} 3\times 12 \\ =36 \end{smallmatrix}}{\mathop{3C}}\,\to \underset{\begin{smallmatrix} 4\times 27 \\ =108 \end{smallmatrix}}{\mathop{4\,Al}}\,+3C{{O}_{2}}\]
$$\because $$  For $$108 g$$  of $$Al,$$ $$36 g$$  of $$C$$ is required in above reaction.
$$\therefore $$  For $$270 \times {10^3}g$$   of $$Al$$  required amount of $$C$$
$$\eqalign{ & = \frac{{36}}{{108}} \times 270 \times {10^3} \cr & = 90 \times {10^3}g \cr & = 90\,kg \cr} $$