Numericals of Chemistry

8/10/2019 Numericals of Chemistry

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Numerical Question Bank

Compiled by:- Dr. Sujata Patil, Chemistry Department, MSSCET, Jalna. Page 1

Matsyodari Shikshan Sansthas

College of Engineering & Technology, Jalna.

Department of Applied Science

Subject: - Engineering Chemistry

1. Water Technology

1. A sample of water on analysis has been found to contain following in ppm,

Sr. No. Impurity Quantity in ppm

1. Ca(HCO3)2 4.86

2. CaSO4 6.80

3. MgSO4 8.40

Calculate the temporary and permanent hardness of the water.

(Ans. 3 ppm and 12 ppm)

2. A sample of water in analysis shows following impurities expressed in mg/lit.

Sr. No. Impurity Quantity in ppm

1. Ca(HCO3)2 220

2. Mg(HCO3)2 56

3. MgCl2 130

4. MgSO4 84

5. CaSO4 98

Calculate amount of lime and soda required to soften 106 liters of water.

(Ans. Lime = 339.6 Kg and Soda = 293.5 Kg)

3. Calculate amount of lime (88.3 % pure) and soda (99.2 % pure) required to soften 24000 liters of water

per day for a year containing the following

CaCO3= 1.85 mg / l CaSO4= 0.34 mg / l

MgCO3= 0.42 mg / l MgCl2= 0.76 mg / l

MgSO4= 0.90 mg / l NaCl = 2.34 mg / l

SiO2= 2.32 mg / l


4. Calculate quantity of lime and soda required for softening 60000 liters of water containing following:-

CO2= 20 mg / l Ca (HCO3)2 = 20 mg / l

Mg (HCO3)2 = 25 mg / l HCl = 8.4 mg / l

Al2(SO4)3= 40 mg / l MgCl2= 12 mg / l



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5. 50 ml of a water sample in Mohrs method of titration with AgNO3 (0.025 M) requires 4.3 ml

Calculate amount of Cl-ions per liter in water sample. Express amount in terms of CaCO3 equivalent.

(Ans. Amount of Cl-ions per liter = 76.325 mg of per liter,

Cl-ions in terms of CaCO3 equivalent = 107.5 mg of CaCO3 equivalent)

6.

A zeolite softener was completely exhausted and was regenerated by passing 100 liter of NaClcontaining 120 gm / liter of NaCl. How many liters of a sample of water of hardness 500 ppm can be

soften by this softener?

(Ans. Volume of water sample = 20512.8 liters)

7. A water sample contains 204 mg of CaSO4per liter. Calculate hardness in terms of CaCO3 equivalent.

(Ans. Hardness = 150 ppm)

2. Chemical Energy Sources

8. Calculate higher and lower calorific value of a coal sample having C = 84 %, S = 1.5 %, N = 0.6 %, H

= 5.5 %, O = 8.4 %.

(Ans. GCV = 8356.05 Kcal / Kg, LCV = 8065.485 Kcal / Kg)

9. Calculate gross and net calorific value of a sample of coal having composition,

C = 80 %, H = 7 %, O = 3 %, S = 3.5 %, N = 2.1 %, ash = 4.4 %.

(Ans. GCV = 8828.025 Kcal / Kg, LCV = 8458.215 Kcal / Kg)

10. Liquid fuel weighing 0.98 g and containing C = 90.1 %, H = 8 %, have following bomb calorimeter

analysis:-Amount of H2O in calorimeter = 1450 g

Water equivalent of calorimeter = 450 g

Rise in temperature of H2O = 1.80C

Latent heat of steam = 587 cal / g

Calculate gross and net calorific value of a sample of coal.

(Ans. GCV = 3489.79 cal / g, LCV = 3067.15 cal / g)

11. 0.945 g of fuel on combustion in excess of oxygen increases temperature of water in a calorimeter

from 13.25 to 19.20C. The mass of water in calorimeter was found to be 1458 gm. Calculate gross

calorific value of fuel, if water equivalent of calorimeter is 144 gm.

(Ans.GCV = 10086.67 cal / g)

12.Temperature of 950 gm of water increased from 25.5 to 28.50C, on burning 0.75 gm of a solid fuel in

a bomb calorimeter. Water equivalent of calorimeter and latent heat of steam are 400 gm and 587 cal /

g respectively. If fuel contains 0.65 % of hydrogen, calculate its net calorific value.

(Ans.GCV = 5400 cal / g, LCV = -5365.66 cal / g)


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774 seconds at 1000F. High viscosity standard (Pennylsylvanian Oil) gave Saybolt viscosity values of

64 seconds at 2100F and 41 seconds at 100

0F. Calculate viscosity index of oil sample under test.

(Ans. Viscosity Index = 58.33)

20. An oil of unknown viscosity index has a Saybolt universal viscosity of 58 seconds at 2100F and of

580 seconds at 100 0F. High viscosity standard (Pennylsylvanian Oil) gave Saybolt viscosity values of58 seconds at 210

0F and 430 seconds at 100

0F. Low viscosity standard (Gulf Oil) possesses a Saybolt

viscosity of 58 seconds at 2100F and 780 seconds at 100

0F. So, calculate viscosity index of oil.


21. A lubricating oil has Saybolt universal viscosity of 59 seconds at 2100F and 564 seconds at 100

0F

Low viscosity standard (Gulf Oil) possesses a Saybolt viscosity of 59 seconds at 2100F and 758

seconds at 1000F. High viscosity standard (Pennylsylvanian Oil) gave Saybolt viscosity values of 59

seconds at 2100F and 420 seconds at 100

0F. Find viscosity index of sample oil.


22. An oil sample has a Saybolt universal viscosity same as that of standard Gulf oil (low V. I.) and

Pennylsylvanian oil (high V. I.) at 2100F. Their Saybolt universal viscosities at 100

0F are 61, 758 and

420 seconds respectively. Calculate viscosity index of oil.


6. Electrochemical Energy Systems

23. 0.01 N KCl solution shows a resistance 225 in a cell. Specific conductivity of 0.1 N KCl at

temperature of experiment is 0.00141 cm-2. If 0.02 N solution shows 80 in same cell, find specific

and equivalent conductance of solution.

(Ans. Specific conductance = 3.966 10-3

ohm-1

cm-1

, Equivalent conductance = 198.28 ohm-1

cm2eq

-1)

24.The resistance of a decinormal solution of a salt occupying a volume between two platinum electrodes

1.80 cm apart and 5.4 cm2

in area was found to be 32 . Calculate equivalent conductance of solution.

(Ans. Equivalent conductance = 104 ohm-1

cm2eq

-1)

25.The resistance of a conductivity cell filled with 0.1 N NaCl solution is 210 at 18 0C. Calculate

equivalent conductance of the solution.(Given: - Cell Constant = 0.88 cm-1

)

(Ans. Equivalent conductance = 41.9 ohm-1

cm2eq

-1)

26. Resistance of a cell with 0.02 M KCl solution at 250C is 225 . The specific conductance of 0.02 M

KCl solution at this temperature was 0.00277 ohm-1

cm-1

. Cell was then and filled with M/10 solution

of CuSO4 and its resistance was found to be 37.5 . Calculate equivalent and molar conductance of

CuSO4 solution.

(Ans. Equivalent conductance = 83 ohm-1

cm2eq

-1, Molar conductance = 166 ohm

-1cm

2mol

-1)

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