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

$$\eqalign{ & 25 \times N = 0.1 \times 35;N = 0.14 \cr & Ba{\left( {OH} \right)_{\text{2}}}\,{\text{is}}\,\,{\text{diacid}}\,{\text{base}} \cr & {\text{hence}}\,N = M \times 2\,\,{\text{or}}\,M = \frac{N}{2} \Rightarrow M = 0.07M \cr} $$

$${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$$
In the given reaction, 1 mol of $${N_2}$$  requires 3 moles of $${H_2}$$  for the formation of ammonia.
Thus if the number of moles of $${N_2}$$  and $${H_2}$$  should be in ration 1 : 3.
We know,
$$n = \frac{{{\text{weight}}}}{{{\text{molecular}}\,{\text{weight}}}}$$
Thus in option A $$56\,g$$  of $${N_2}$$  means 2 moles of $${N_2}$$  and $$10\,g$$  of $${H_2}$$  means 5 moles of $${H_2}.$$
Thus here $${H_2}$$  acts as limiting reagent.

At $$NTP$$  22400 $$cc$$  of $${N_2}O$$  contains $$ = 6.02 \times {10^{23}}{\text{molecules}}$$
$$\therefore \,\,1\,cc\,{N_2}O$$   will contain $$ = \frac{{6.02 \times {{10}^{23}}}}{{22400}}{\text{molecules}}$$
In $${N_2}O$$  molecule, number of atoms = 2 + 1 = 3
Thus, number of atoms
$$ = \frac{{3 \times 6.02 \times {{10}^{23}}}}{{22400}}{\text{atoms}}$$
$$ = \frac{{1.8 \times {{10}^{22}}}}{{224}}{\text{atoms}}$$
In $${N_2}O$$  molecule, number of electrons = 7 + 7 + 8 = 22
Hence, number of electrons
$$ = \frac{{6.02 \times {{10}^{23}}}}{{22400}} \times 22{\text{ electrons}}$$
$$ = \frac{{1.32 \times {{10}^{23}}}}{{224}}{\text{ electrons}}$$

$$\eqalign{ & N{a_2}C{O_3}\,\,\,\,\,\,\,\,\,\,NaHC{O_3} \cr & x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 - x} \right) \cr & \frac{x}{{106}} = \frac{{1 - x}}{{84}}\,{\text{given (moles are equal )}} \cr & x = 0.557 \cr & \frac{{0.557}}{{53}} + \frac{{0.443}}{{84}} = \frac{{V \times 0.1}}{{1000}} \cr & V = 157.7\,mL \cr} $$

$${\text{Number of moles}} \propto \frac{1}{{{\text{Molecular mass}}}}$$
Molecular mass of $$C{O_2} = 44,{N_2} = 28,C{H_4} = 16,HCl = 36.5$$
$$C{O_2}$$  will have least volume, as no. of moles is directly proportional to volume at constant $$P$$  and $$T.$$

\[\begin{align} & \text{The given equations are} \\ & MgC{{O}_{3\left( s \right)}}\xrightarrow{\Delta }Mg{{O}_{\left( s \right)}}+C{{O}_{2\left( g \right)}} \\ & CaC{{O}_{3\left( s \right)}}\xrightarrow{\Delta }Ca{{O}_{\left( s \right)}}+C{{O}_{2\left( g \right)}} \\ \end{align}\]
$${\text{Let total mass of the mixture}} = 1.0\,g$$
$${\text{Then mass of}}\,\,MgC{O_3} = x\,g$$       $${\text{and mass of }}CaC{O_3} = 1 - x\,g$$
$$\eqalign{ & {\text{According to the above equation,}} \cr & 84\,\,g\,\,MgC{O_3}\,\,{\text{gives}}\,\,44\,\,g\,\,C{O_2} \cr & \therefore \,\,x\,\,g\,\,MgC{O_3}\,\,{\text{gives}}\,\frac{{44}}{{84}}x\,\,g\,\,C{O_2} \cr & {\text{Again}}\,\,100\,\,g\,\,CaC{O_3}\,\,{\text{gives}}\,\,44\,\,g\,\,C{O_2} \cr} $$
$$\therefore \,\,\left( {1 - x} \right)g\,\,CaC{O_3}\,\,{\text{gives}}$$      $$\frac{{44}}{{100}}\left( {1 - x} \right)g\,\,C{O_2}$$
$${\text{According to the problem}},$$     $$\frac{{44x}}{{84}} + \frac{{44}}{{100}}\left( {1 - x} \right) = 0.5\,g$$
$${\text{On solving we get,}}\,x = 0.715\,g$$
$${\text{Hence, percentage of}}$$     $$MgC{O_3} = 0.715 \times 100 = 71.5\% $$
$${\text{and percentage of}}$$     $$CaC{O_3} = 100 - 71.5 = 28.5\% $$

Molar mass of
$$\eqalign{ & CuS{O_4} = 63.5 + 32 + 4 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 159.5\,g \cr} $$
Mass of copper present in $$159.5\,g$$   of $$CuS{O_4} = 63.5\,g$$
∴ Mass of copper present in $$50\,g$$  of $$CuS{O_4}$$
$$\eqalign{ & = \frac{{63.5}}{{159.5}} \times 50 \cr & = 19.90\,g \cr} $$

$$\eqalign{ & 1\,mole = 6.023 \times {10^{23}}\,molecule \cr & 18\,g = 6.02 \times {10^{23}}molecule \cr} $$
$$18\,g = {\text{mass}}\,{\text{of}}\,6.02 \times {10^{23}}$$      $${\text{water}}\,{\text{molecules}}$$
$${\text{Mass}}\,{\text{of}}\,{\text{one}}\,{\text{water}}\,{\text{molecule}}$$      $$ = \frac{{18}}{{6.023 \times {{10}^{23}}}}g$$
$$\eqalign{ & {\text{Density}} = 1\,g\,c{m^{ - 3}} \cr & {\text{Volume}} = \frac{{{\text{Mass of one water molecule}}}}{{{\text{Density}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{6.023 \times {{10}^{23}} \times 1}}c{m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \approx 3.0 \times {10^{ - 23}}c{m^3} \cr} $$

$$6CaO + {P_4}{O_{10}} \to 2C{a_3}{\left( {P{O_4}} \right)_2}$$
$$1\,mole$$  of $${P_4}{O_{10}}$$  $$=$$ molar mass of $${P_4}{O_{10}} = 284\,g$$
$$852\,g$$  of $${P_4}{O_{10}} = \frac{{852}}{{284}} = 3\,mol$$
$$1\,mole$$  of $${P_4}{O_{10}}$$  reacts with $$6\,moles$$   of $$CaO$$
$$3\,moles$$  of $${P_4}{O_{10}}$$  reacts with $$18\,moles$$   of $$CaO$$
Mass of $$18\,moles$$   of $$CaO = 18 \times 56 = 1008\,g$$

\[\underset{1\,g-eq}{\mathop{M{{g}^{2+}}}}\,+\underset{1\,g-eq}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,\to MgC{{O}_{3}}+2N{{a}^{+}}\]
$$\eqalign{ & 1\,g{\text{ - equivalent of}} \cr & M{g^{2 + }} = 12g\,\,{\text{of}}\,M{g^{2 + }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12000\,mg\,{\text{of}}\,M{g^{2 + }} \cr} $$
$${\text{Now,}}\,12000\,mg\,\,{\text{of}}\,M{g^{2 + }} \equiv $$       $$1000\,{\text{milliequivalent of}}\,N{a_2}C{O_3}$$
$$\therefore \,\,12\,mg\,\,{\text{of}}\,M{g^{2 + }} \equiv $$     $$1\,{\text{milliequivalent of}}\,N{a_2}C{O_3}$$