Solution for the mathematical Problems

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 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
Now putting these values in the equation (i) we get,                             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

Now putting these values in the equation (i) we get,                             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 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=?

                                                 

Determination of Molality

Now putting these values in the equation (i) we get,                             So , Molality of the final solution is 1.52474m

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=?

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.
Now putting these values in the equation (i) we get,                             So , Molality of the final solution is 0.100 m	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=?

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.
Now putting these values in the equation (i) we get,                             So , Normality  of the final solution is 0.200 N	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=?

Problem 5. [11(a); Page 551]  49 g of H2SO4 are dissolved in 250 mL of solution. Calculate the molarity of the solution.
Now putting these values in the equation (i) we get,                             So , Molarity of the final solution is 2.00M	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=?

Problem 6. [11(b); Page 551]  45 g of glucose, C6H12O6, are dissolved in 500 g of water.Calculate molality of the solution.
Now putting these values in the equation (i) we get,                             So , Molality of the final solution is 0.500m	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=?

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.

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.

Determination of Molarity

Now putting these values in the equation (i) we get,                             So , Molarity of the final solution is 0.0847 M

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=?

Determination of Molality

Now putting these values in the equation (i) we get,                             So , Molality of the final solution is 0.08547 m

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=?

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.

Determination of Molarity

Now putting these values in the equation (i) we get,                             So , Molarity of the final solution is 0.050 M

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=?

Determination of Normality

Now putting these values in the equation (i) we get,                             So , Normality of the final solution is 0.10 N

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=?

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

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.
Now putting these values in the equation (i) we get,                             So , Molarity of the final solution is 4.534 M	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=?

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?
Now putting these values in the equation (i) we get,                             So , Normality of the final solution is 1.2478 N	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=?

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, 

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 .
Now putting these values in the equation (i) we get,                             So , Molality of the final solution is 0.40 m	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=?

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 =13 (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

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
Now putting these values in the equation (i) we get,                             So , The amount of acid needed is 0.490 gm	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

Problem 18. [34 ; Page 556]  How many grams of glucose are present in 100 mL of 0.1 M solution
Now putting these values in the equation (i) we get,                             So , The amount of glucose needed is 1.800 g	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

Problem 19. [38 ; Page 557]  49 g of H2SO4 is dissolved in 250 mLof the solution, find out the  molarity of the solution .
Now putting these values in the equation (i) we get,                             So , Molarity of the final solution is 2.00M	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=?
Problem 20. [42 ; Page 557]  What is the total weight of 100 ml of 2 M solution of HCl?
Now putting these values in the equation (i) we get,                             So , The  total amount =(100+7.3)gm=107.3 gm	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?

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

Problem 22. [48 ; Page 558]  What is the weight of urea required to prepare 200 ml of 2 M solution?
Now putting these values in the equation (i) we get,                             So , The weight of urea required=24.0 gm	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

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?
Now putting these values in the equation (i) we get,                             So , Molality of the final solution is 0.200 m	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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