JAMB Chemistry Past Questions & Answers - Page 312

1,556.

2.0g of a monobasic acid was made up to 250 cm3 with distilled water. 25.00 cm3 of this solution required 20.00cm3 of 0.1 M NaOH solution for complete neutralization. The molar mass of the acid is?

A.

200 g

B.

160 g

C.

100 g

D.

50 g

Correct answer is C

Ma = (MbVb)/(Va) = (0.1 m χ 20 cm3)/(25cm) = 0.08 m

M = (concentration in g/dm3)/(m) = (8)/(0.08) = 100 g

1,557.

Na2C2O4 + CaCl2 → CaC2O4 + 2NaCl. Given a solution of 1.9g of sodium oxalate in 50g of water at room temperature, calculate the minimum volume of 0.1 M calcium oxalate using the above equation?

A.

1.40 χ 102 dm3

B.

14.0 χ 102 cm3

C.

1.40 χ 10-2 dm-2

D.

14.0 χ 10-2 cm3

Correct answer is A

Na2C2O4 + CaCl2 → CaC2O4 + 2NaCl

molar mass of Na2C2O4 = 134

molarity in 1000g H2O = (1.9)/(134) = 0.014 m

but in 502 g mole = (0.014)/(50) χ 1000 = 0.28 m

M1V1 = M2V2
V2 = (M1V1)/(M2) = (0.28 χ 50)/(0.1) = 140
= 1.40 χ 102 dm3

1,558.

Smoke consist of?

A.

solid particles dispersed in liquid

B.

solid or liquid particles dispersed in gas

C.

gas or liquid particles dispersed inliquid

D.

liquid particles dispersed inliquid

Correct answer is B

No explanation has been provided for this answer.

1,559.

The solubility in moles per dm3 of 20g of CuSO4 dissolve in 100 of water at 180°C is?

(Cu = 63.5, S = 32, O = 16)

A.

0.13

B.

0.25

C.

1.25

D.

2.00

Correct answer is C

Solubility in moles per dm3

No of moles = (20)/(159.5) χ 0.125 moles

in 100 g of moles = (0.125)/(100) χ 1000 = 1.25 moles

1,560.

A blue solid, T, which weighed 5.0g was placed on a table. After 8 hours, the resulting pink solid was found to weigh 5.5g. It can be inferred that substance T?

A.

is deliquescent

B.

is hydroscopic

C.

has some molecules of water of crystallization

D.

is efflorescent

Correct answer is A

No explanation has been provided for this answer.