WAEC Past Questions and Answers - Page 1430

7,146.

The pressure exerted by a gas is a function of the

A.

total volume of the gas

B.

speed of the gaseous molecules

C.

mass of each gaseous molecule

D.

frequency of collision between gaseous molecules

Correct answer is D

No explanation has been provided for this answer.

7,147.

Which of the following phenomenna lead to decrease in volume of a liquid in an open container?

A.

Brownian motion

B.

Diffusion

C.

Evaporation

D.

Sublimation

Correct answer is C

No explanation has been provided for this answer.

7,148.

The gas law which describes the relationship between volume and temperature is

A.

Boyle's law

B.

Charles' law

C.

Dalton's law

D.

Graham's law

Correct answer is B

No explanation has been provided for this answer.

7,149.

The percentage by mass of calcium in Ca(OCI)2 is [Ca = 40.0; CI = 35.5; O = 16.0]

A.

28.0%

B.

31.6%

C.

43.8%

D.

44.5%

Correct answer is A

Molar mass of \(Ca(OCl)_2\) = 40 + (16 + 35.5)x2 = 143

% by mass of calcium in the compound =\(\frac{mass  of  calcuim}{molar  mass  of  compound}\) x 100%

% calcium = \(\frac{40}{143} \times100\)

  = 27.98% = 28%

7,150.

The volume of 0.25 moldm-3 solution of KOH that would yield 6.5g of of solid KOH on evaporation is (K = 39.0; o = 16.0; H = 1.00)

A.

464.30 cm3

B.

625.00 cm3

C.

1000.00 cm3

D.

2153.80 cm3

Correct answer is A

number of moles = molar concentration x volume(\(dm^3\))

molar concentration(C) = 0.25mol/dm3, volume = ? and number of mole(n) = \(\frac{mass}{molar mass}\) = \(\frac{6.5}{56}\) ( where 56g/mol = molar mass of KOH)

\(\frac{6.5}{56}\) = 0.25 x V 

V = \(\frac{0.11607}{0.25}\) = 0.46428dm^3 → 464.3cm^3