Engineering Questions with Answers - Multiple Choice Questions

# Applied Chemistry MCQ – Problems, Units and Formulae

1 - Question

Hardness in water is expressed in terms of ____________ equivalents.
a) Calcium carbonate
b) Calcium bicarbonate
c) Magnesium hydroxide
d) Magnesium oxide

Explanation: Hardness of the water is expressed in terms of the calcium carbonate equivalents. They are like ppm, degree Clark and French unit.

2 - Question

1 degree Clark is equal to ________ ppm.
a) 12.3
b) 13.3
c) 14.3
d) 15.3

Explanation: One degree Clark is equal to the 14.3 ppm. Ppm means parts per million. 1ppm is equal to 1mg/litre.

3 - Question

One French unit is equal to __________ ppm.
a) 10
b) 20
c) 30
d) 40

Explanation: One French unit is equal to the 10ppm. Hardness causing salt as the number of parts of the substance by weight in million parts by weight of water is ppm.

4 - Question

The ppm is one part of calcium carbonate equivalent hardness is present in __________ of water.
a) One
b) One million
c) One billion
d) One trillion

Explanation: The ppm is one part of calcium carbonate equivalent hardness is present in the one million parts of water.

5 - Question

One French unit is equal to _________ mg/litre.
a) 5
b) 10
c) 15
d) 20

Explanation: One French unit is equal to 10mg/litre. One French unit is equal to 10ppm and 0.7 degree Clark.

6 - Question

50 ml of standard and hard water containing 1mg of pure CaCO3 per ml consumed 10ml of EDTA solution. 50ml of given EDTA sample requires 10ml of same EDTA solution. Calculate the total hardness of water sample in ppm.
a) 10ppm
b) 100ppm
c) 1000ppm
d) 10000ppm

Explanation: 50ml of standard hard water requires 10ml of EDTA solution so 1ml of standard water requires 5ml of EDTA solution. So, 50ml of water sample requires 10ml of EDTA solution. So, 50ml of water sample requires 50mg of CaCO3. So, 1000ml of water sample requires 50*(1000/50)=1000 mg of CaCO3 that is 1000ppm.

7 - Question

In determination of hardness by EDTA method, 50ml of standard hard water required 30ml of EDTA solution while 50ml of sample hard water consumed 20ml of EDTA solution. After boiling 50ml of same sample required 10ml of EDTA solution. Calculate the permanent hardness.
a) 322ppm
b) 332ppm
c) 664ppm
d) 644ppm

Explanation: For 50ml of boiled water requires the 10ml of EDTA solution that is 10*(50/30) mg of CaCO3. The 1000ml of the boiled water requires the 10*(50/30)*20=322mg of CaCO3. It means permanent hardness is 322ppm.

8 - Question

We know that lime required for softening of water is x{temp Ca hardness + 2.Mg hardness + perm(Mg+Fe+3Al)hardness + 1/2 HCL + H2SO4 – NaAlO2 – CO2}-all are in terms of mg of CaCO3. Here x=?
a) 7.4
b) 0.74
c) 74
d) 740

Explanation: Lime also reacts with bicarbonates of Na and K to form carbonate. Since 100 parts of CaCO3 is equivalent to the 74 parts of Ca(OH)2. so, x=74/100=0.74.

9 - Question

100 parts of CaCO3 is equivalent to the __________ parts of sodium carbonate.
a) 103
b) 104
c) 105
d) 106

Explanation: 100 parts of CaCO3 is equivalent to the 106 parts of the sodium carbonate. So, the washing soda requirement is 100/106{temp Ca hardness+2.Mg hardness+perm(Mg+Fe+3Al)hardness+1/2 HCL+H2SO4-NaAlO2-CO2}

10 - Question

The chemical oxygen demand can be given as __________
a) {[(V1-V2)*N*8]}/x
b) {[(V1+V2)*N*8]}/x
c) {[(V2-V1)*N*8]}/x
d) {[(V1/V2)*N*8]}/x

Explanation: The chemical oxygen demand can be given by {[(V1-V2)*N*8]}/x where V1=volume of ferrous ammonium sulphate required for blank, V2=volume of ferrous ammonium sulphate required for test, N=normality of ferrous ammonium sulphate, x=volume of sewage sample taken.

11 - Question

The biochemical oxygen demand can be given by ___________
a) (Dob – Dos)+dilution factor
b) (Dob – Dos)-dilution factor
c) (Dob – Dos)/dilution factor
d) (Dob – Dos)*dilution factor

Explanation: The biochemical oxygen demand can be given by (Dob – Dos)*dilution factor where Dob = dissolved oxygen present in blank Dos = dissolved oxygen of sewage after incubation.

12 - Question

If a sample water has not supplied any heat and having impurities as follows: Mg(HCO3)2=50 mg of CaCO3, MgSO4 = 100mg of CaCO3, CaCl2=200mg of CaCO3, Ca(NO3)2=100mg of CaCO3. Calculate the lime required for treatment of 10000 litres of water.
a) 1.82Kg
b) 1.50Kg
c) 1.45Kg
d) 1.48Kg