NEET Chemistry Solved Question Paper 2011

1 mole of H₂SO₄ = 2g-equivalent of H₂SO₄ .Hence , when 1 mole of H₂SO₄ is mixed with 2 moles of NaOH , the heat evolved will be 2 x 57.3 kJ .

${}_{27}{}^{60}\mathrm{Co}$ is radioactive and unstable due to high $\frac{\mathrm{n}}{\mathrm{p}}$ ratio .

$\left(\mathrm{i}.\mathrm{e}. , \frac{\mathrm{n}}{\mathrm{p}} > 1\right)$

Number of protons = 27

Number of neutrons = 33

$\therefore$ $\frac{\mathrm{n}}{\mathrm{p}}$ for ${}_{27}{}^{60}\mathrm{Co}$ = $\frac{33}{27}$ = 1.22

Entropy is the measure of degree of disorder (or randomness) of a system . The greater the disorder in a system , the higher is the entropy . Hence , entropy is highest for hydrogen gas .

Conjugate acid is obtained , when a base accepts a proton .

$\underset{\mathrm{base}}{{\mathrm{CH}}_3{\mathrm{NH}}_2}$ + $\underset{\mathrm{proton}}{{\mathrm{H}}^{+}}$ $\to$$\underset{\mathrm{conjugate} \mathrm{acid}}{{\mathrm{CH}}_3{\mathrm{NH}}_3^{+}}$

$∆$m = 0.090 u

Number of nucleons = 9

$\therefore$ Binding energy per nucleon

= $\frac{\mathrm{total} \mathrm{binding} \mathrm{energy}}{\mathrm{number} \mathrm{of} \mathrm{nucleon}}$

= $\frac{∆\mathrm{m} \times 931.5}{\mathrm{number} \mathrm{of} \mathrm{nucleon}}$

= $\frac{0.090 \times 931.5}{9}$ = 9.315 MeV

The substance which. can donate a pair of electrons is called Lewis base .

Amines contain a lone pair of electrons on nitrogen atom which is available for donation , so they behave as Lewis base .

$\mathrm{\Delta }$S = 2.303nR log $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

n = 1

$\therefore$ $∆$S = 2.303 R log $\frac{10}{1}$

$∆$S = 2.303R              (log 10 = 1)

Normality of the mixed solution 

= $\frac{{\mathrm{N}}_1{\mathrm{V}}_1 + {\mathrm{N}}_2{\mathrm{V}}_2}{{\mathrm{V}}_1 + {\mathrm{V}}_2}$

= $\frac{0.6 \times \frac{100}{1000} + 0.3 \times {\displaystyle \frac{200}{1000}}}{{\displaystyle \frac{100 + 200}{1000}}}$

= $\frac{0.6 \times 0.1 \times 0.3 \times 0.2}{0.3}$

= $\frac{0.06 + 0.06}{0.3}$ 

= $\frac{0.12}{0.3}$ = 0.4 N

N₂ + 3H₂ $\rightleftharpoons$ 2NH₃

$∆$H = $∆$E + $∆$n_gRT

$∆$n_g = n_p - n_r = 2 - 4 = -2

$\therefore$ $∆$H = $∆$E + (-2)RT

$∆$H = $∆$E - 2RT

50 mL of0.1 M HCl = $\frac{0.1 \times 50}{1000}$ = 5 x 10^-3

50 mL of 0.2 M NaOH = $\frac{0.2 \times 50}{1000}$ = 10 x 10^-3

Hence , after neutralisation NaOH left 

= 10 x 10^-3 - 5 x 10^-3 = 5 x 10^-3

Total volume = 100 cc

The concentration of NaOH

= $\frac{5 \mathrm{x} {10}^{-3} \mathrm{x} 1000}{100}$ = 0.05 M

[OH^-] = 0.05 M = 5 x 10^-2 M

pOH = - log [OH^-]

= - log [5 x 10^-2]

= 1.3010

pH + pOH = 14

pH = 14 - 1.3010

= 12.699

= 12.70