This set of Applied Chemistry Problems focuses on “Problems, Units and Formulae”.
1. Hardness in water is expressed in terms of ____________ equivalents.
a) Calcium carbonate
b) Calcium bicarbonate
c) Magnesium hydroxide
d) Magnesium oxide
View Answer
Explanation: Hardness of the water is expressed in terms of the calcium carbonate equivalents. They are like ppm, degree Clark and French unit.
2. 1 degree Clark is equal to ________ ppm.
a) 12.3
b) 13.3
c) 14.3
d) 15.3
View Answer
Explanation: One degree Clark is equal to the 14.3 ppm. Ppm means parts per million. 1ppm is equal to 1mg/litre.
3. One French unit is equal to __________ ppm.
a) 10
b) 20
c) 30
d) 40
View Answer
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. 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
View Answer
Explanation: The ppm is one part of calcium carbonate equivalent hardness is present in the one million parts of water.
5. One French unit is equal to _________ mg/litre.
a) 5
b) 10
c) 15
d) 20
View Answer
Explanation: One French unit is equal to 10mg/litre. One French unit is equal to 10ppm and 0.7 degree Clark.
6. 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
View Answer
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. 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
View Answer
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. 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
View Answer
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. 100 parts of CaCO3 is equivalent to the __________ parts of sodium carbonate.
a) 103
b) 104
c) 105
d) 106
View Answer
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. 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
View Answer
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. 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
View Answer
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. 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
View Answer
Explanation: Lime needed is 0.74[temp Ca hardness+temp of Mg hardness+perm Mg hardness]. So, 0.74(0+2*50+100)=148mg/litre. Lime requirement for 10000 litres of water is 148*10000mg=1.48Kg.
Sanfoundry Global Education & Learning Series – Applied Chemistry.
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