JAMB Chemistry Past Questions & Answers - Page 261

1,301.

A given amount of gas occupies 10.0 dm3 at 4atm and 273°C. The number of moles of the gas present is?
(Molar volume of a gas at s.t.p. = 22.4 dm3)

A.

0.89 mol

B.

1.90 mol

C.

3.80 mol

D.

5.70 mol

Correct answer is A

\(V_1\) = \(10dm^3\), \(V_2\) = ?, \(P_1\) = 4atm =  4 x 760 = 3040mmHg, \(P_2\) = 760mmHg \(T_1\) = 273ºc = 273 + 273 = 546K, \(T_2\) = 273K

Using the general gas law = \(\frac{P_1 V_1}{ T_1} = \frac{P_2 V_2}{T_2}\)

\(V_2  = \frac{P_1 V_1 T_2}{P_2 T_1}\)

\(V_2 = \frac{ 3040 \times10 \times273}{760 \times546}\)

\(V_2 = 20dm^{-3}\)

1mole of a gas = \(22. 4dm^{3}\)

  x                     = \(20.0dm^{3}\)

= 0.89mol.

1,302.

A mixture of iron and sulphur can be separated by dissolving the mixture in?

A.

steam

B.

dilute hydrochloric acid

C.

dilute sodium hydroxide

D.

benzene

Correct answer is D

No explanation has been provided for this answer.

1,303.

The addition of water to calcium oxide leads to?

A.

a physical change

B.

a chemical change

C.

the formation of a mixture

D.

an endothermic change

Correct answer is B

No explanation has been provided for this answer.

1,304.

(1 /(2) N2(g) + (1) / (2)O2(g) → NO(g) ∆H° 89 KJ mol-. If the entropy change for the reaction above at 25°C is 11.8 J mol-, calculate the change in free energy. ∆G°, for the reaction at 25°C?

A.

88.71 KJ

B.

85.48 KJ

C.

-204.00 KJ

D.

-3427.40 KJ

Correct answer is D

∆G = ∆H - T∆S; T = 2s° + 273 = 298 k
∆G = 89 - 298 x 11.8 = -3427.4

1,305.

25 cm3 of 0.02 M KOH neutralized 0.03 g of a monobasic organic acid having the general formula CnH2n + 1 COOH. The molecular formula of the acid is?

(C = 12, H = 1, O = 16)

A.

HCOOH

B.

C2H5COOH

C.

CH3COOH

D.

C3H7COOH

Correct answer is C

Conc. of KOH = 39 x 0.02 = 0.78 g/dm3
25cm3 require 0.03 g of CnH2n + 1 COOH 1000CM3 → x x = (1000) / (25) x 0.03 = 1.2g
CnH2n + 1COOH + KOH → CnH2n + 1 COOK + H2O
from (MAVA) / (MB x VB) = (1) / (1) (MA x 1.2) / (0.02 x 0.78) = (1) / (1)
MA = 0.013
Con. = Molarity x Molecular Mass
Molecular Mass = (0.78) / (0.013) x 60
CnH2n + 1 COOH = 60
12n + 1 (2n + 1) + 12 + 32 + 1 = 6
14n = 60 - 46; n = (14) / (14) = 1
∴ CH3COOH