Chemistry MCQs for NEET — Practice Questions with Answers

$$\begin{gathered} {\mathrm{I}}_2 + 2{\mathrm{Cl}}_2 \stackrel{}{\to } \mathrm{ICI} + {\mathrm{ICI}}_3 \\ \mathrm{Mole} \mathrm{before} \mathrm{reaction} \frac{25.4}{254} \frac{14.2}{71} 0 0 \\ =0.1 =0.2 0 0 \\ \mathrm{Mole} \mathrm{after} \mathrm{reaction} 0 0 0.1 0.1 \\ \left(\mathrm{Note}: \mathrm{Mole} \mathrm{ratio} \mathrm{of} \mathrm{reaction}\right) \\ \mathrm{Mole} \mathrm{of} \mathrm{ICI} \mathrm{formed} = 0.1 \\ \mathrm{Mole} \mathrm{of} {\mathrm{ICI}}_3 \mathrm{formed} = 0.1 \end{gathered}$$

$$\begin{gathered} \mathrm{Given} \mathrm{vapour} \mathrm{density} \mathrm{of} \mathrm{mixture} = 38.3 \\ \therefore \mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{mixture} = 2\times 38.3 = 76.6 \\ \mathrm{Let} \mathrm{a} \mathrm{g} \mathrm{of} {\mathrm{NO}}_2 \mathrm{be} \mathrm{present} \mathrm{in} 100 \mathrm{g} \mathrm{mixture}, \\ \therefore \mathrm{g} \mathrm{of} {\mathrm{N}}_2{\mathrm{O}}_4 \mathrm{in} \mathrm{mixture} = \left(100-\mathrm{a}\right) \\ \mathrm{Therefore}, \frac{\mathrm{a}}{46}+\frac{100-\mathrm{a}}{92}=\frac{100}{76.6} \\ \therefore \mathrm{a} = 20.10 \mathrm{g} \\ \therefore \mathrm{Mole} \mathrm{of} {\mathrm{NO}}_2 \mathrm{in} \mathrm{mixture} = 20.10/46 \\ = 0.437 \end{gathered}$$

Let valencies of Cu in two oxides be x and y, then

           I oxides is ${\mathrm{Cu}}_2{\mathrm{O}}_{\mathrm{x}}$

           II oxides is ${\mathrm{Cu}}_2{\mathrm{O}}_{\mathrm{y}}$

In I oxide:

            Equivalent of Cu = Equivalent of oxygen

                             $\frac{\mathrm{w}}{\mathrm{A}/\mathrm{x}}=\frac{\mathrm{a}}{8} ....\left(\mathrm{i}\right)$

where w, A, x and a are mass of Cu, at. mass of Cu, valency of Cu and mass of oxygen

In II oxide:

                              $\frac{\mathrm{w}}{\mathrm{A}/\mathrm{y}}=\frac{\mathrm{a}}{8} ....\left(\mathrm{ii}\right)$

By Eqs. (i) and (ii)

                             $\frac{\mathrm{x}}{\mathrm{y}}=\frac{2}{1}$

$\therefore$ Valency of Cu in I and II oxides are in the ratio 2:1

Suppose formula of hydrocarbon is ${\mathrm{C}}_{\mathrm{a}}{\mathrm{H}}_{\mathrm{b}}$

$\therefore {\mathrm{C}}_{\mathrm{a}}{\mathrm{H}}_{\mathrm{b}}+\left[\mathrm{a}+\left(\mathrm{b}/4\right)\right]{\mathrm{O}}_2 \stackrel{}{\to } {\mathrm{aCO}}_2 + \left(\mathrm{b}/2\right){\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

Volume taken initially

      $5 \mathrm{mL} 30 \mathrm{mL} - -$

Volume after reaction

      $0 30-5\left[\mathrm{a}+\mathrm{b}/4\right] 5\mathrm{a} -$

$\because$ Given that, ${\mathrm{O}}_2$ left is absorbed by pyrogallol

Also volume absorbed by NaOH = 10 mL = Volume of ${\mathrm{CO}}_2$ formed

$\therefore 5\mathrm{a}=10 \therefore \mathrm{a}=2$

Total volume of gases left after reaction = 25 mL

or Volume of ${\mathrm{O}}_2$ left + Volume of ${\mathrm{CO}}_2$ formed = 25 mL

$\therefore$   Volume of ${\mathrm{O}}_2$ left = 25-10 =15

$\therefore$   30-5 [a+(b/4)]=15   $\therefore$ b=4

Thus, hydrocarbon is ${\mathrm{C}}_2{\mathrm{H}}_4$.

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{mixture} \mathrm{contains} \mathrm{a} \mathrm{g} {\mathrm{CaCO}}_3 \mathrm{and} \mathrm{b} \mathrm{g} {\mathrm{MgCO}}_3. \\ \therefore \mathrm{a}+\mathrm{b}=2.21 .....\left(\mathrm{i}\right) \\ \mathrm{On} \mathrm{heating} {\mathrm{CaCO}}_3 \stackrel{}{\to } \mathrm{CaO}+{\mathrm{CO}}_2 \\ {\mathrm{MgCO}}_3 \stackrel{}{\to } \mathrm{MgO}+{\mathrm{CO}}_2 \\ \because \mathrm{Molar} \mathrm{of} \mathrm{CaO}=\mathrm{mole} \mathrm{of} {\mathrm{CaCO}}_3=\mathrm{a}/100 \\ \therefore \mathrm{Mass} \mathrm{of} \mathrm{CaO}=\left(\mathrm{a}/100\right)\times 56 \mathrm{g} \\ \mathrm{Similarly}, \mathrm{mass} \mathrm{of} \mathrm{MgO}=\left(\mathrm{b}/84\right)\times 40 \mathrm{g} \\ \mathrm{Therefore}, \frac{56\mathrm{a}}{100}+\frac{40\mathrm{b}}{84}=1.152 .....\left(\mathrm{ii}\right) \\ \left(\mathrm{i}.\mathrm{e}., \mathrm{mass} \mathrm{of} \mathrm{residue} \mathrm{left}\right) \\ \mathrm{Solving} \mathrm{Eqs}. \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \mathrm{a}=1.19 \mathrm{g}; \mathrm{b}=1.02 \mathrm{g} \\ \mathrm{Also}, \mathrm{mole} \mathrm{of} {\mathrm{CO}}_2 \mathrm{formed} = \mathrm{mole} \mathrm{of} {\mathrm{CaCO}}_3+\mathrm{mole} \mathrm{of} {\mathrm{MgCO}}_3 \\ = \frac{1.19}{100}+\frac{1.02}{84}=0.0241 \\ \therefore \mathrm{Volume} \mathrm{of} {\mathrm{CO}}_2 \mathrm{at} \mathrm{NTP} = 0.0241\times 22400 \\ = 539.8 \mathrm{mL} \end{gathered}$$

$$\begin{gathered} \mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{metal} \mathrm{chloride}=95\times 2=190 \mathrm{g} {\mathrm{mol}}^{-1} \\ \mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{metal}=\frac{6.4}{0.13}=49.23 \mathrm{g} {\mathrm{mol}}^{-1} \\ \mathrm{Let} \mathrm{the} \mathrm{metal} \mathrm{chloride} \mathrm{be} {\mathrm{MCl}}_{\mathrm{n}} \\ \mathrm{Then} 49.23+\mathrm{n}\times 35.5=190 \\ \therefore \mathrm{n}=3.9\approx 4; \\ \mathrm{Thus} \mathrm{exact} \mathrm{atomic} \mathrm{mass}\left(\mathrm{a}\right) \mathrm{of} \mathrm{metal} \mathrm{can} \mathrm{be} \mathrm{derived} \mathrm{by} \\ \mathrm{a}+4\times 35.5 = 190 \\ \therefore \mathrm{a} = 48 \\ \therefore \mathrm{E} = \frac{48}{4} \\ = 12 \mathrm{g} {\mathrm{eq}}^{-1} \end{gathered}$$

Humid air ${\mathrm{N}}_2+{\mathrm{O}}_2+{\mathrm{H}}_2{\mathrm{O}}_{\mathrm{v}}$ is lighter than dry air $\left({\mathrm{N}}_2+{\mathrm{O}}_2\right)$ as its average molar mass is less. Presence of $\mathrm{He}$ in air $\left({\mathrm{N}}_2+{\mathrm{O}}_2\right)$ gives lower molar mass than humid air. $\mathrm{He}$ has lower molar mass than ${\mathrm{H}}_2{\mathrm{O}}_{\mathrm{v}}$.

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{one} \mathrm{molecule} \mathrm{of} {\mathrm{C}}_{60} {\mathrm{H}}_{122}=\left(12\times 60\right)+122 \mathrm{amu} \\ = 842 \mathrm{amu} \\ = 842\times 1.66\times {10}^{-24} \mathrm{g} \\ = 1.4\times {10}^{-21} \mathrm{g} \end{gathered}$$

Explanation is correct reason for statement.

ans 3,

Certain elements combine with other atoms, donating, accepting or sharing electrons in different proportions depending on the nature of the reaction.