NEET Practice Questions (MCQs) with Answers & Solutions

(A).   Permutite is sodium aluminium silicate. The ${\mathrm{Na}}^{+}$ ions are ionisable. So, during use, these

         ${\mathrm{Na}}^{+}$ ions are exchanged with ${\mathrm{Ca}}^{2+}, {\mathrm{Mg}}^{2+},$ etc. present in hard water.

${\mathrm{Na}}^{+} {\left(\mathrm{Al}-\mathrm{silicate}\right)}^{–} +{\mathrm{Ca}}^{2+} \left(\mathrm{aq}\right) \stackrel{ }{\to }\mathrm{Ca} {\left(\mathrm{Al}-\mathrm{silicate}\right)}_2 + 2{\mathrm{Na}}^{+} \left(\mathrm{aq}\right)$

(B).  Due to small atomic size of aluminium, the size of vacancy in the lattice structure is also

        small, so it cannot absorb hydrogen. 

(C).  The IE of hydrogen is much higher than those of alkali metals and slightly higher than those

        of halogens. For example IE of Cl is 1255 kJ/mole and IE of H is 1312 kJ/mole. 

(D).   It is colourless gas

(A). In metallic hydride, hydrogen atoms are occupied the vacancies in between metal atoms.

(C).  ${}_1{}^2\mathrm{H}$ represents deuterium which consists of one ${\mathrm{e}}^{–}$, one p and one n.

(C). These are the properties of ${\mathrm{H}}_2{\mathrm{O}}_2$.

$$\begin{gathered} \left(\mathrm{c}\right) 25.1 \mathrm{ml} \mathrm{of} 0.02\left(\mathrm{N}\right){\mathrm{H}}_2{\mathrm{SO}}_4=25.1\times 0.02\times {10}^{-3} \mathrm{equiv}. \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4 \\ \mathrm{Now}, 25.1\times 0.02\times {10}^{-3} \mathrm{equiv}. \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4=25.1\times 0.02\times {10}^{-3} \\ \mathrm{equiv}. \mathrm{of} {\mathrm{CaCO}}_3=25.1\times 0.02\times {10}^{-3}\times 50 \mathrm{gm} \mathrm{of} {\mathrm{CaCO}}_3 \\ \therefore \mathrm{Hardness} \mathrm{of} \mathrm{given} \mathrm{water} \mathrm{sample} = 25.1\times 0.02\times {10}^{-5}\times 50\times {10}^6=251 \mathrm{ppm}. \end{gathered}$$

(C). A mixture of ${\mathrm{H}}_2{\mathrm{O}}_2$ and hydrazine $\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$ is used as a rocket fuel.

$$\begin{gathered} \left(\mathrm{B}\right). \mathrm{The} \mathrm{wt}. \mathrm{of} 50 \mathrm{lit}. \mathrm{of} \mathrm{hard} \mathrm{water}=50\times {10}^3\mathrm{gm} \\ \therefore \mathrm{wt}. \mathrm{of} {\mathrm{CaCO}}_3 \mathrm{present}=\frac{120\times 50\times {10}^3}{{10}^6}\mathrm{gm} \\ 50 \mathrm{gm} {\mathrm{CaCO}}_3=81 \mathrm{gm} \mathrm{of} \mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2 \\ \therefore \frac{120\times 50\times {10}^3}{{10}^6}\mathrm{gm} {\mathrm{CaCO}}_3 \\ \equiv \frac{81}{50}\times \frac{120\times 50\times {10}^3}{{10}^6}\mathrm{gm} \mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2 \\ \equiv 81\times 120\times {10}^3 \mathrm{gm} \mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2 \\ \equiv 9.72 \mathrm{gm} \mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2 \\ \equiv 0.06 \mathrm{moles} \mathrm{of} \mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2 \\ \therefore \mathrm{moles} \mathrm{of} \mathrm{Ca}{\left(\mathrm{OH}\right)}_2 \mathrm{added}=\frac{1.62\times {10}^3}{40}=40.5 \\ \therefore \mathrm{moles} \mathrm{Ca}{\left(\mathrm{OH}\right)}_2 \mathrm{reacted} =0.06 \\ \mathrm{moles} \mathrm{Ca}{\left(\mathrm{OH}\right)}_2 \mathrm{remaining} =40.5-0.06=40.44 \\ \therefore \mathrm{moles} {\mathrm{Ca}}^{2+} \mathrm{in} 50 \mathrm{lit}.=40.44 \\ \therefore \mathrm{moles} {\mathrm{CaCO}}_3 \mathrm{in} 50 \mathrm{lit}.=40.44 \\ \therefore \mathrm{amount} \mathrm{of} {\mathrm{CaCO}}_3 \mathrm{in} 50 \mathrm{lit}.=40.44\times 100=4044 \mathrm{gm} \\ \therefore \mathrm{hardness} \mathrm{or} \mathrm{resulting} \mathrm{water} \\ =\frac{4044}{5\times {10}^4}\times {10}^6=8.088\times {10}^4 \mathrm{ppm} \end{gathered}$$