Solution to the Mathematical Problems from Bahl Tuli MANIK

Solution to the 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

Solution of the Mathematical Problem
Prepared By: Md. Imran Nur Manik, Lecturer, Department of Pharmacy, NUB Page 1
manikrupharmacy@gmail.com www.pharmacydocs.blogspot.com
Solution for the mathematical Problems
Chapter: Solution
Book: Essentials of physical chemistry (Multicolor Edn. 2009)
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
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



Now putting these values in the equation (i)
we get,
0.755447SOr,
1081.6105.9
100086.53
S




So , Molality of the final solution is 0.75447 m.
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=?
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.
Determination of Molarity
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
4459.1SOr,
743.9198
100013
S




So , Molarity of the final solution is 1.4459 M
Here,
The amount of solute, W=13 gm
Molecular weight of solute, M=98 g/mol
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=?
Md.Imran
Nur
Manik

Determination of Molality
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
52474.1SOr,
8798
100013
S




So , Molality of the final solution is 1.52474m
Here,
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
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.
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
100.0SOr,
5040
10000.2
S




So , Molality of the final solution is 0.100 m
Here,
The amount of solute, W=0. 2 gm
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.
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
200.0SOr,
50063
10006.3
S




So , Normality of the final solution is 0.200 N
Here,
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=?
Md.Imran
Nur
Manik

Problem 5. [11(a); Page 551] 49 g of H2SO4 are dissolved in 250 mL of solution. Calculate the molarity
of the solution.
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
00.2SOr,
25098
100049
S




So , Molarity of the final solution is 2.00M
Here,
Molecular weight of solute , M=98 g/mol
Problem 6. [11(b); Page 551] 45 g of glucose, C6H12O6, are dissolved in 500 g of water.Calculate
molality of the solution.
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
500.0SOr,
500180
100045
S




So , Molality of the final solution is 0.500m
Here,
Molecular weight of solute , M=180 g/mol
Problem 7. [25(b); Page 552] A sample of spirit
contains 92% of ethanol by weight
the rest being water. What is the mole fraction of
its constituents?
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.
Md.Imran
Nur
Manik

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.
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
08478.0SOr,
024.100858.5
10005
S




Here,
Molecular weight of solute, M=58.5g/mol
Total amount of solution=(1000+5) gm=1005 gm
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
Determination of Molality
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
08547.0SOr,
100058.5
10005
S




Here,
Molecular weight of solute, M=58.5 g/mol
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.)
Md.Imran
Nur
Manik

Problem 9. [29 ; Page 552] Calculate the molarity and normality of a solution containing 5.3 g of
Na2CO3 dissolved in 1000 mL solution.
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
050.0SOr,
1000106
10005.3
S




Here,
Determination of Normality
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
10.0SOr,
100053
10005.3
S




Here,
Equivalent weight of the solute, M=(106÷2) gm
=53 gm
The volume of the final solvent, V = 1000 mL
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.
Solution: We know, Molecular weight of NaCl= 58.5 gm Where, M.Wt. of Na+
= 23g & Cl–
= 35.5g
17.25gm
1000
500231.5
WOr,
1000
SMV
Wsolution,M1.5inionsNaofamountNow, +




Therefore the amount of Na+
ions in grams present=17.25 gm
Md.Imran
Nur
Manik

26.625gm
1000
50035.51.5
WOr,
1000
SMV
Wsolution,M1.5inionsClofamountNow, -




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.
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
534.4SOr,
200036.5
1000331
S




Here,
Molecular weight of solute , M=36.5 g/mol
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?
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
2478.1SOr,
40056.1
100028
S




Here,
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
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.
Solution:
Here,
Molecular weight of solute, M=56.1 gm/mol
Molarity of the Solution, S=6.90 M
The volume of the final solution, V = ?
).........(..........
SM
1000W
VOr,
1000
SMV
Wthat,knowWe
i



Md.Imran
Nur
Manik

Now putting these values in the equation (i) we get,
5013.77VOr,
1.566.9
100030
V




Therefore the volume of the solution= 77.5013 mL
And the total amount of solution= 70+30= 100gm.
gm/mL2903.1
77.0513
100
Or,
solutionfinalftheovolume
graminsolutionofmass
Density,Now



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 .
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
40.0SOr,
500180
100036
S




Here,
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.
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 (ans.)
Problem 16. [32 ; Page 556] Calculate the molarities of 0.1N solution of HCl and 0.1N solution of
H2SO4.
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
Md.Imran
Nur
Manik

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
).......(..........
1000
SMV
Wthat,knowWe i
we get,
490.0WOr,
1000
100490.1
W



So , The amount of acid needed is 0.490 gm
Here,
The amount of solute, W=?
Equivalent weight of the solute, M=(98÷2) gm
=49 gm
Normality of the Solution, S=0.1N
Problem 18. [34 ; Page 556] How many grams of glucose are present in 100 mL of 0.1 M solution
).......(..........
1000
SMV
Wthat,knowWe i
we get,
800.1WOr,
1000
1001800.1
W



So , The amount of glucose needed is 1.800 g
Here,
Molecular weight of solute =180g/mol
Molarity of the Solution, S=0.1M
Problem 19. [38 ; Page 557] 49 g of H2SO4 is dissolved in 250 mLof the solution, find out the molarity
of the solution .
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
00.2SOr,
25098
100049
S




So , Molarity of the final solution is 2.00M
Here,
Molecular weight of solute, M=98g/mol
Md.Imran
Nur
Manik

Problem 20. [42 ; Page 557] What is the total weight of 100 ml of 2 M solution of HCl?
).......(..........
1000
SMV
Wthat,knowWe i
we get,
3.7WOr,
1000
1005.362
W



So , The total amount =(100+7.3)gm=107.3 gm
Here,
Molecular weight of solute =36.5 g/mol
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?
Solution: ).......(..........
solutionofamountTotal
soluteofAmount
ionConcentratthat,knowWe i
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
Problem 22. [48 ; Page 558] What is the weight of urea required to prepare 200 ml of 2 M solution?
).......(..........
1000
SMV
Wthat,knowWe i
we get,
24WOr,
1000
200602
W



So , The weight of urea required=24.0 gm
Here,
Molarity of the Solution, S=2 M
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?
).........(..........
MV
1000W
SOr,
1000
SMV
Wthat,knowWe
i



we get,
200.0SOr,
50092
10009.2
S




Here,
Md.Imran
Nur
Manik

Solution to the Mathematical Problems from Bahl Tuli MANIK

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Solution to the Mathematical Problems from Bahl Tuli MANIK

Similar to Solution to the Mathematical Problems from Bahl Tuli MANIK (20)

More from Imran Nur Manik

More from Imran Nur Manik (20)

Recently uploaded

Recently uploaded (20)

Solution to the Mathematical Problems from Bahl Tuli MANIK