Answer to Question #143793 in General Chemistry for Kelechi Agbi Jr

Question

Molarity/Normality:
Determine the molarity of the following solutions:
31. 84 gms of NaHCO3 in 250 mL of solution
32. 6.0 gms of NaOH in 1.0 L of solution
33. 25 gms of NaCl in 500 mL of solution
34. 35 gms of H2SO4 in 450 mL of solution
How many grams are necessary to make:
35. 300 mL of 0.20 M NaCl
36. 75 mL of 0.750 M KCl
37. 250 mL of 4 M MgSO4
Determine the normality of the following solutions:
38. 42 gms of K3PO4 in 500 mL of solution
39. 245 gms of NaOH in 1L of solution
40. 20 gms of H3PO4 in 300 mL of solution

Answer 1

When, W grams of solute having Molar mass M grams present in V ml of solution.

Then the formula of Molarity of the solution,

     C = (W×1000)/(M×V) M

31.

Molar mass of NaHCO3, M = 84 g/mol

Mass of NaHCO3 dissolve, W = 84 gms

Volume of the solution, V = 250 mL

C = (W×1000)/(M×V) M

    = (84×1000)/(84×250) M

    = 4 M

    

    So, molarity of the solution =4 M

32.

Molar mass of NaOH, M = 40.00 g/mol

Mass of NaOH dissolve, W = 6.0 gms

Volume of the solution, V = 1.0 L

                                             = 1000 mL

C = (W×1000)/(M×V) M

   = (6×1000)/(40×1000) M

   = 0.15 M

   

 So, molarity of the solution =0.15 M

33.

Molar mass of NaCl, M = 58.44 g/mol

Mass of NsCl dissolve, W = 25 gms

Volume of the solution, V = 500 mL

C = (W×1000)/(M×V) M

   = (25×1000)/(58.44×500) M

   =0.855 M

   

So, molarity of the solution = 0.855 M

   

34.

Molar mass of H2SO4, M = 98.00 g/mol

Mass of H2SO4 dissolve, W = 35 gms

Volume of the solution, V = 450 mL

C = (W×1000)/(M×V) M

  = (35×1000)/(98×450) M

  =0.8274 M

  

So, molarity of the solution = 0.8274 M

  

35.

Molar mass of NaCl, M = 58.44 g/mol

Molarity of the solution, C = 0.20 M

Volume of the solution, V =300 mL

      C = (W×1000)/(M×V)

Or, W = (M×V)×C/1000

Or, W = ( 58.44×300)×0.2/1000

         = 3.51 g

         

So, 3.51g NaCl is necessary to make the above solution

36.

Molar mass of KCl, M = 74.5513 g/mol

Molarity of the solution, C = 0.750 M

Volume of the solution, V =75 mL

      C = (W×1000)/(M×V)

Or, W = (M×V)×C/1000

Or, W = (74.5513 ×75)×0.750/1000

         =4.2 g

         

So, 4.2 g of KCl is necessary to make the above solution

37.

Molar mass of MgSO4, M = 120.366 g/mol

Molarity of the solution, C = 4 M

Volume of the solution, V = 250 mL

     C = (W×1000)/(M×V)

Or, W = (M×V)×C/1000

Or, W = (120.366×250)×4/1000

         =120.366g

         

So, 120.366 g of MgSO4 is necessary to make the above solution

When, W grams of solute having Molar mass M grams present in V ml of solution.

Then the formula of normarity of the solution,

     C = (W×1000)/(E×V) N

     

38.

Equivalent mass of K3PO4,

E = 106.135 g/eqv.

Mass of K3PO4 present in solution, W= 42 gms

Volume of the solution, V = 500 mL

      C = (W×1000)/(E×V) N

 Or,C = (42×1000)/(106.135×500) N

         = 0.79 N

         

So, normality of the solution = 0.79 N

39.

Equivalent mass of NaOH, E = 20.00 g/eqv.

Mass of NaOH present in solution,W = 245 gms

Volume of the solution, V = 1L

                                             = 1000 mL

       C = (W×1000)/(E×V) N

 Or, C = (245×1000)/(20×1000) N

          = 12.25 N

          

So, normality of the solution = 12.25 N

40.

Equivalent mass of H3PO4, E =48.997 g/eqv.

Mass of H3PO4 present in solution, W = 20 gms

Volume of the solution, V =300 mL

       C = (W×1000)/(E×V) N

 Or, C =(20×1000)/(48.997×300) N

           = 1.361 N

So, normality of the solution =1.361 N