Solution to the
Mathematical
Problems
Mathematical
Problems
Chapter: Solution
Book: Essentials of physical chemistry
(Multicolor Edn. 2009)
Prepared By
Md. Imran Nur Manik
Lecturer
Department of Pharmacy
Northern University Bangladesh
Book: Essentials of physical chemistry
(Multicolor Edn. 2009)
Prepared By
Md. Imran Nur Manik
Lecturer
Department of Pharmacy
Northern University Bangladesh
Solution for the mathematical Problems
Chapter: Solution
Chapter: Solution
Book: Essentials of physical chemistry (Multicolor Edn. 2009)
By-Md. Imran Nur Manik, Lecturer, Pharmacy, NUB
By-Md. Imran Nur Manik, Lecturer, Pharmacy, NUB
Using the formula: W=(SMV÷1000)
Problem 1. [5(b); Page 551] Determine the molality of a solution containing 86.53 g of sodium carbonate (mol mass = 105.99 gm) per litre in water at 20°C. The density of the solution at this temperature is 1.0816 g mL–1
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 Now putting these values in the equation (i) we get, 
So , Molality of the final solution is 0.75447 m.
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Here,
The amount of solute, W=86.53 gm
Molecular weight of solute, M=105.9 g/mol
Density of the solution, D= 1.0816 gm/mL
That means, 1 mL solution = 1.0816 gm
So, 1L solution = (1.0816× 1000) gm
= 1081.6 gm.
Therefore,
The amount of the final solvent, V = 1081.6 gm
Molality of the Solution, S=?
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Problem 2. [7(a) ; Page 551] What is molarity and molality of a 13% solution (by weight) of H2SO4. It’s density is 1.09 g/mL.
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Determination of Molarity
Now putting these values in the equation (i) we get, 
So , Molarity of the final solution is 1.4459 M
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Here,
The amount of solute, W=13 gm
Molecular weight of solute, M=98 g/mol
Density of the solution, D= 1.09 gm/mL
That means, 1.09 gm solution = 1mL
So, 100 gm solution = (100÷1.09) mL
= 91.743 mL
Therefore,
The volume of the final solution, V = 91.743 mL
Molarity of the Solution, S=?
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Determination of Molality
Now putting these values in the equation (i) we get, 
So , Molality of the final solution is 1.52474m
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Here,
The amount of solute, W=13 gm
Molecular weight of solute, M=98 g/mol
The solution is 13% by weight.
That means, the amount of solvent = (100÷13) gm
= 87 gm
Therefore,
The amount of the final solvent, V = 87 gm
Molality of the Solution, S=?
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Problem 3. [8 ; Page 551] Calculate the molality of a solution of sodium hydroxide which contains 0.2 g of sodium hydroxide in 50 g of the solvent.
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Now putting these values in the equation (i) we get, 
So , Molality of the final solution is 0.100 m
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Here,
The amount of solute, W=0. 2 gm
Molecular weight of solute, M=40 g/mol
The amount of the final solvent, V = 50 gm
Molality of the Solution, S=?
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Problem 4. [9 ; Page 551] Calculate the normality of a solution containing 6.3 g of oxalic acid crystals (Mol. wt. 126) dissolved in 500 mL of solution.
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Now putting these values in the equation (i) we get, 
So , Normality of the final solution is 0.200 N
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Here,
The amount of solute, W=6.3 gm
Molecular weight of solute =126 g/mol
Equvalent weight of solute, M=(126÷2) gm
=63 gm
The volume of the final solution, V = 500 mL
Normality of the Solution, S=?
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Problem 5. [11(a); Page 551] 49 g of H2SO4 are dissolved in 250 mL of solution. Calculate the molarity of the solution.
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Now putting these values in the equation (i) we get, 
So , Molarity of the final solution is 2.00M
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Here,
The amount of solute, W=49 gm
Molecular weight of solute , M=98 g/mol
The volume of the final solution, V = 250 mL
Molarity of the Solution, S=?
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Problem 6. [11(b); Page 551] 45 g of glucose, C6H12O6, are dissolved in 500 g of water.Calculate molality of the solution.
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Now putting these values in the equation (i) we get, 
So , Molality of the final solution is 0.500m
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Here,
The amount of solute, W=45 gm
Molecular weight of solute , M=180 g/mol
The amount of the final solvent, V = 500 gm
Molality of the Solution, S=?
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Problem 7. [25(b); Page 552] A sample of spirit contains 92% of ethanol by weight
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the rest being water. What is the mole fraction of its constituents?
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Solution: Let the mass of solution be 100 gm
Since it contains 92% of ethanol by weight so the wight of ethanol is 92 gm and that of water is 8 gm.
Now , no. of moles of ethanol=(92 ÷ 46) mole =2 mole
no. of moles of water = (8 ÷ 18) mole = 0.444 mole
Total moles of solution= (2+0.444) mole = 2.444 mole
Therefore, the mole fraction of ethanol= (2÷2.444) = 0.818
And the mole fraction of water = (0.444÷2.444) = 0.1816~0.182
Mole fraction of ethanol= 0.818 and water= 0.182.
Problem 8. [27 ; Page 552] 5 g of NaCl is dissolved in 1 kg of water. If the density of the solution is 0.997 g mL–1, calculate the molarity, normality, molality and mole fraction of the solute.
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Determination of Molarity
Now putting these values in the equation (i) we get, 
So , Molarity of the final solution is 0.0847 M
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Here,
The amount of solute, W=5 gm
Molecular weight of solute, M=58.5g/mol
Total amount of solution=(1000+5) gm=1005 gm
Density of the solution, D= 0.997 gm/mL
That means, 0.997 gm solution = 1mL
So, 1005 gm solution = (100÷1.09) mL
= 1008.024072 mL
Therefore,
The volume of the final solution, V = 1008.024 mL
Molarity of the Solution, S=?
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Determination of Molality
Now putting these values in the equation (i) we get, 
So , Molality of the final solution is 0.08547 m
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Here,
The amount of solute, W=5 gm
Molecular weight of solute, M=58.5 g/mol
The amount of the final solvent, V = 1000 gm
Molality of the Solution, S=?
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Determination of Normality
We know N=nM ; Thus ,N=1×0.0847 (Where n is the equivalent number of NaCl=1
And, M is the Molarity of the solution= 0.0847 , Known earlier )
So, Normality of the solution=0.0847 N
Determination of Mole Fraction
Here no. of moles of solute, NaCl=(5÷58.5)mole = 0.0855 mole
And no. of moles of Solvent, H2O = (1000÷18) mole = 55.5 mole
Total moles of solute and solvent = (0.0855+55.5) mole = 55.5855 mol.
Mole fraction of NaCl = (0.0855÷ 55.5855) =0.001538. (ans.)
Problem 9. [29 ; Page 552] Calculate the molarity and normality of a solution containing 5.3 g of Na2CO3 dissolved in 1000 mL solution.
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Determination of Molarity
Now putting these values in the equation (i) we get, 
So , Molarity of the final solution is 0.050 M
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Here,
The amount of solute, W=5.3 gm
Molecular weight of solute, M=106 g/mol
The volume of the final solution, V = 1000 mL
Molarity of the Solution, S=?
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Determination of Normality
Now putting these values in the equation (i) we get, 
So , Normality of the final solution is 0.10 N
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Here,
The amount of solute, W=5 gm
Molecular weight of solute =106 g/mol
Equivalent weight of the solute, M=(106÷2) gm
=53 gm
The volume of the final solvent, V = 1000 mL
Normality of the Solution, S=?
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Determination of Normality from Molarity
We know N=nM ; Thus ,N=2×0.05 (Where n is the equivalent number of Na2CO3=2
And, M is the Molarity of the solution= 0.05 , Known earlier )
So, Normality of the solution=0. 10 N
Problem 10. [28 ; Page 552] Calculate the amount of Na+ and Cl– ions in grams present in 500 mL of 1.5 M NaCl solution.
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Solution: We know, Molecular weight of NaCl= 58.5 gm Where, M.Wt. of Na+ = 23g & Cl–= 35.5g
Therefore the amount of Na+ ions in grams present=17.25 gm
Therefore the amount of Cl– ions in grams present=26.625 gm
Problem 11. [30; Page 552] Calculate the molarity of a solution containing 331g of HCl dissolved in sufficient water to makes 2dm3 of solution.
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Now putting these values in the equation (i) we get, 
So , Molarity of the final solution is 4.534 M
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Here,
The amount of solute, W=331 gm
Molecular weight of solute , M=36.5 g/mol
The volume of the final solvent, V = 2000 mL
Molality of the Solution, S=?
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Problem 12. [33; Page 552] What is the normality of a solution containing 28.0 g of KOH dissolved in sufficient water to make 400 ml of solution?
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Now putting these values in the equation (i) we get, 
So , Normality of the final solution is 1.2478 N
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Here,
The amount of solute, W=28 gm
Molecular weight of solute =56.1 g/mol
Equivalent weight of the solute, M=(56.1÷1) gm
=56.1 gm
The volume of the final solvent, V = 400mL
Normality of the Solution, S=?
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Problem 13. [34 ; Page 552] A 6.90 M solution of KOH in water contains 30% by weight of KOH. Calculate the density of the solution.
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Solution:
Here,
The amount of solute, W=30 gm
Molecular weight of solute, M=56.1 gm/mol
Molarity of the Solution, S=6.90 M
The volume of the final solution, V = ?
Now putting these values in the equation (i) we get,
Therefore the volume of the solution= 77.5013 mL
And the total amount of solution= 70+30= 100gm.
So the density of the solution=1.29 gm/mL
Problem 14. [28 ; Page 556] 36 g of glucose (molecular mass 180) is present in 500 g of water, find out the molality of the solution .
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Now putting these values in the equation (i) we get, 
So , Molality of the final solution is 0.40 m
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Here,
The amount of solute, W=36 gm
Molecular weight of solute, M=180 g/mol
The amount of the final solvent, V = 500 gm
Molality of the Solution, S=?
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Problem 15. [30 ; Page 556] Find out the mole fraction of ethyl alcohol in a solution containing 36 g of H2O and 46 g of ethyl alcohol.
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Solution: Here, moles of water = (36÷18) mole = 2 moles
Moles of ethyl alcohol = (46÷46) mole =1 mole
Total moles of the solution = (2+1) mole= 3 moles
So, mole fraction of ethyl alcohol =13 (ans.)
Problem 16. [32 ; Page 556] Calculate the molarities of 0.1N solution of HCl and 0.1N solution of H2SO4.
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Solution: We know N=nM
Here, N= Normality of the final solution, M= Molarity of the final solution and n= Equivalent number of the solute
Determination of Molarity of 0.1N solution of HCl
We know N=nM ; Thus ,0.1=1×M (Where n is the equivalent number of HCl=1
And, N is the Normality of the solution= 0.1 , Known earlier )
So, Molarity of the solution=0. 10 M
Determination of Molarity of 0.1N solution of H2SO4
We know N=nM ; Thus ,0.1=2×M (Where n is the equivalent number of H2SO4=2
And, N is the Normality of the solution= 0.1 , Known earlier )
So, Molarity of the solution=0. 05 M
Problem 17. [32 ; Page 556] Find out the amount required for the preparation of 100ml of 0.1N H2SO4
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 Now putting these values in the equation (i) we get, 
So , The amount of acid needed is 0.490 gm
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Here,
The amount of solute, W=?
Molecular weight of solute =98 g/mol
Equivalent weight of the solute, M=(98÷2) gm
=49 gm
The volume of the final solvent, V = 100mL
Normality of the Solution, S=0.1N
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Problem 18. [34 ; Page 556] How many grams of glucose are present in 100 mL of 0.1 M solution
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Now putting these values in the equation (i) we get, 
So , The amount of glucose needed is 1.800 g
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Here,
The amount of solute, W=?
Molecular weight of solute =180g/mol
The volume of the final solvent, V = 100mL
Molarity of the Solution, S=0.1M
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Problem 19. [38 ; Page 557] 49 g of H2SO4 is dissolved in 250 mLof the solution, find out the molarity of the solution .
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 Now putting these values in the equation (i) we get, 
So , Molarity of the final solution is 2.00M
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Here,
The amount of solute, W=49 gm
Molecular weight of solute, M=98g/mol
The volume of the final solvent, V = 250 mL
Molarity of the Solution, S=?
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Problem 20. [42 ; Page 557] What is the total weight of 100 ml of 2 M solution of HCl?
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Now putting these values in the equation (i) we get, 
So , The total amount =(100+7.3)gm=107.3 gm
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Here,
The amount of solute, W=?
Molecular weight of solute =36.5 g/mol
The volume of the final solvent, V = 100mL
Molarity of the Solution, S=2 M
The weight of the final solvent = 100 gm |
Problem 21. [43 ; Page 557] 1 kg of a solution of CaCO3 contains 1 g of calcium carbonate. What will be the concentration of the solution?
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Solution:
Here, the amount of solute=1 gm and the total amount of solution =1000 gm
Now putting these values in the equation (i) we get, (1÷1000)=0.001
In terms of ppm the concentration will be =(0.001× 106)= 1000 ppm
In terms of ppm the concentration will be =(0.001× 106)= 1000 ppm
Problem 22. [48 ; Page 558] What is the weight of urea required to prepare 200 ml of 2 M solution?
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Now putting these values in the equation (i) we get, 
So , The weight of urea required=24.0 gm
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Here,
The amount of solute, W=?
Molecular weight of solute =60 g/mol
The volume of the final solvent, V = 200 mL
Molarity of the Solution, S=2 M
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Problem 23. [49 ; Page 558] What is the molality of a solution prepared by dissolving 9.2 g toluene (C7H8) in 500 g of benzene?
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Now putting these values in the equation (i) we get, 
So , Molality of the final solution is 0.200 m
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Here,
The amount of solute, W=9.2 gm
Molecular weight of solute, M=92 g/mol
The amount of the final solvent, V = 500 gm
Molality of the Solution, S=?
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