Some Basic Concepts in Chemistry MCQs for NEET — Chemistry Questions with Answers

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

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. 

Let suppose we use x–litre water gas at S.T.P.

So it contain 0.5x litre H₂.

Now according to given equation

CO   +  H₂O            $\stackrel{}{\to }$ CO₂  +   H₂

0.45x  Excess              0.05x    0.5x  (45% CO , and 5% CO2)

0  –(0.05x + 0.45x)    (0.5x + 0.45x)

So total volume of H₂ = 0.95x

But according to question

It is 5 litre

So 0.95x = 5

x = 5.263 litre