1

### AIPMT 2001

Solubility of M2S salt is 3.5 $\times$ 10$-$6 then find out solubility product.
A
1.7 $\times$ 10$-$6
B
1.7 $\times$ 10$-$16
C
1.7 $\times$ 10$-$18
D
1.7 $\times$ 10$-$12

## Explanation

M2S ⇌ 2M+ + S2–

Ksp = [M+]2[S2–] = (2s)2(s) = 4s 3

Ksp = 4(3.5 × 10–6)3 = 1.7 × 10–16
2

### AIPMT 2001

Ionisation constant of CH3COOH is 1.7 $\times$ 10$-$5 and concentration of H+ ions is 3.4 $\times$ 10$-$4. Then find out initial concentration of CH3COOH molecules.
A
3.4 $\times$ 10$-$4
B
3.4 $\times$ 10$-$3
C
6.8 $\times$ 10$-$4
D
6.8 $\times$ 10$-$3

## Explanation

CH3COOH ⇌ CH3COO + H+

Kc = ${{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}}$

$\Rightarrow$ [CH3COOH] = ${{{3.4 \times {{10}^{ - 4}} \times 3.4 \times {{10}^{ - 4}}} \over {1.7 \times {{10}^{ - 5}}}}}$

= 6.8 × 10–3
3

### AIPMT 2001

In HS$-$, I$-$, R $-$ NH2, NH3 order of proton accepting tendency will be
A
I$-$ > NH3 > R $-$ NH2 > HS$-$
B
NH3 > R $-$ NH2 > HS$-$ > I$-$
C
R $-$ NH2 > NH3 > HS$-$ > I$-$
D
HS$-$ > R $-$ NH2 > NH3 > I$-$

## Explanation

Strong base has higher tendency to accept the proton. Increasing order of base and hence the order of accepting tendency of proton is

R $-$ NH2 > NH3 > HS$-$ > I$-$
4

### AIPMT 2000

Equilibrium constant Kp for following reaction
MgCO3(s) $\rightleftharpoons$ MgO(s) + CO2(g)
A
Kp = PCO2
B
Kp = PCO2 $\times$ ${{{P_{C{O_2}}} \times {P_{MgO}}} \over {{P_{MgC{O_3}}}}}$
C
Kp = ${{{P_{C{O_2}}} + {P_{MgO}}} \over {{P_{MgC{O_3}}}}}$
D
Kp = ${{{P_{MgC{O_3}}}} \over {{P_{C{O_2}}} + {P_{MgO}}}}$

## Explanation

Kp = PCO2

As solids do not exert pressure, so their partial pressure is taken as unity.