Chemistry questions and answers

Chemistry Questions and Answers

Learn more about the properties, composition, and structure of substances (elements and compounds) with these Chemistry questions and answers. This Test can be used by students preparing for Chemistry in JAMB, WAEC, NECO or Post UTME.

606.

When substance X was added to a solution of bromine water, the solution became colourless. X is likely to be

A.

propane

B.

propanoic acid

C.

propyne

D.

propanol

Correct answer is C

Thus bromine water will be decolorized by alkynes due to the presence of two pi bonds available for reaction!

607.

When a salt is added to its saturated solution, the salt

A.

dissolves and the solution becomes super saturated

B.

dissolves and the solution becomes unsaturated

C.

precipitates and the solution remains unchanged

D.

dissolves and crystals are formed

Correct answer is C

The point of saturation is the point it gets to and from thence, the solution can no longer take in the salt anymore. From this point, it begins to form precipitate while the solution remains unchanged. 

608.

At what temperature does the solubility of \(KNO_{3}\)  equal that of \(NaNO_{3}\)?   

A.

0°C

B.

20°C

C.

30°C

D.

40°C

Correct answer is B

From the graph, check for their point of intersection and trace it down to the temperature axis.

609.

Pure water can be made to boil at a temperature lower than 100°C by

A.

reducing its quantity

B.

decreasing the external pressure

C.

distilling it

D.

increasing the external pressure

Correct answer is B

The easiest and most drastic way to change the boiling point is by tampering with the pressure above your water. To lower the boiling point, you need simply to decrease its external pressure.

610.

What is the mass of solute in 500\(cm^{3}\) of 0.005\(moldm^{-3}\) \(H_{2}SO_{4}\)? ( H = 1, S = 32.0, O=16.0)

A.

0.490g

B.

0.049g

C.

0.245g

D.

0.0245g

Correct answer is C

Using 

n = cv where n = no of mole, c = molar concentration(mol/dm3) and v = volume( dm3)

volume =  500\(cm^{3}\) = 0.5dm3, c = 0.005\(moldm^{-3}\)

number of mole(n) = c x v = 0.5 X 0.005 = 0.0025mol.

but n = \(\frac{mass}{ molar mass}\) =  0.0025 = \(\frac{mass}{ 98}\) (where 98g/mol is the molar mass of H2SO4)

mass of H2SO4 = 0.0025 x 98 = 0.245g.