Chemistry MCQs for NEET — Practice Questions with Answers

(a) The tendency to gain electron is in the order Z > Y > X Thus, 

$Y + e \stackrel{ }{\to }{Y}^{-}, X \stackrel{ }{\to }{X}^{+} +\hspace{0.17em}e$

(b)

 $$\begin{gathered} {\mathrm{E}}_{\mathrm{RP}} = {\mathrm{E}}_{\mathrm{RP}}^{\circ } + \frac{0.059}{2}\log \left[{\mathrm{Cu}}^{2+}\right]; \\ \mathrm{If} \left[{\mathrm{Cu}}^{2+}\right] = \frac{1}{10} \\ \mathrm{Then} {\mathrm{E}}_{\mathrm{RP}} = {\mathrm{E}}_{\mathrm{RP}}^{\circ } +\hspace{0.17em}\frac{0.059}{2}\log \left(\frac{1}{10}\right) \end{gathered}$$

(c) Mg cannot be obtained by electrolysing the aqueous salt solutions. Also $E_{OP}^{\circ }$ order suggests for Ag, Hg, Cu.

(d) Difluoro acetic acid is strong acid.

we know

Q= i t

Q= $100\times {10}^{-3}\times 30\times 60$                              [ current qlways in Ampere time always in seconds]

Q=180 C

(a) A cation having maximum value of standard reduction potential is the strongest oxidising agent. Hence, Mg²⁺ is the strongest oxidising agent.

(b) By Kohlrausch's law 

 $$\begin{gathered} {\mathrm{\lambda }}_{\infty } \mathrm{for} \mathrm{NaCl} = {\mathrm{\lambda }}_{{\mathrm{Na}}^{+}} + {\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}} ...\left(\mathrm{i}\right) \\ {\mathrm{\lambda }}_{\infty } \mathrm{for} \mathrm{HCl} = {\mathrm{\lambda }}_{\mathrm{H}+} + {\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}} ...\left(\mathrm{ii}\right) \\ {\mathrm{\lambda }}_{\infty } \mathrm{for} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{COONa} = {\mathrm{\lambda }}_{{\mathrm{Na}}^{+}} + {\mathrm{\lambda }}_{{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{COO}}^{-}} ...\left(\mathrm{iii}\right) \\ \mathrm{So}, {\mathrm{\lambda }}_{\infty } \mathrm{for} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{COOH} \\ \mathrm{On} \mathrm{adding} \mathrm{Eqs}. \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right) \mathrm{and} \mathrm{then} \mathrm{subtracting} \\ \mathrm{Eq}. \left(\mathrm{i}\right) \\ ={\mathrm{\lambda }}_{\infty } \mathrm{of} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{COONa} + {\mathrm{\lambda }}_{\infty } \mathrm{of} \mathrm{HCl} - {\mathrm{\lambda }}_{\infty } \mathrm{for} \mathrm{NaCl} \\ =(91 + 426.16 - 126.45){\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2 \\ = 390.71 {\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2 \end{gathered}$$

 Concept used: electrolysis and Faradays' first law

Give: acidulated water

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4\to 2{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2^{-}} \\ {\mathrm{H}}_2\mathrm{O} \rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{O}}^{2-} \end{gathered}$$

at acthode ${\mathrm{H}}_2$ gas evolved

w = 2 x Q

$\mathrm{i}=\left(\frac{\mathrm{w}}{\mathrm{E}}\right)\times \frac{96500}{\mathrm{t}}$

for gas at place of $\frac{\mathrm{w}}{\mathrm{E}}$ we can put

$$\begin{gathered} \frac{\mathrm{vol} \mathrm{of} \mathrm{gas} \mathrm{at} \mathrm{NTP}}{\mathrm{vol} \mathrm{of} 1 \mathrm{gm} \mathrm{eq}. \mathrm{gas}} \\ =\frac{112}{{\displaystyle \frac{22400}{2}}} \\ =\frac{112}{11200} \\ =\frac{1}{100} \\ \mathrm{i}=\left(\frac{1}{100}\right) \times \frac{96500}{965} \\ \mathrm{i}=1\mathrm{A} \end{gathered}$$

(a) Degree of dissociation, $\mathrm{\alpha }$ = ${\mathrm{\lambda }}^{\mathrm{c}}$/${\mathrm{\lambda }}^{\infty }$

where, ${\mathrm{\lambda }}^{\mathrm{c}}$and ${\mathrm{\lambda }}^{\infty }$ are equivalent conductances at a given concentration and at infinite dilution respectively. 

= $\mathrm{\alpha }$ = 8/400 = 2x10^-2

From Ostwald's dilution law (for weak monobasic acid),

K_c = C$\mathrm{\alpha }$²/(1-$\mathrm{\alpha }$)

C$\mathrm{\alpha }$²

= 1/32 (2x10^-2)²

=1.25x10^-6

Explanation: charge on 1 mol electron

Q= ne

Q= 6.002 x ${10}^{23} \times 1.6\times {10}^{-19}$

Q = 96500 = 1F

to deposite 1g Eq. of 1 sub 1F electricity which is equal to charge on 1 mole of electron is required.