Calculate the normality and the molarity of the solution given. Solute density,

Question

Calculate the normality and the molarity of the solution given.

             Solute                                                     density, g/mL. Wt.-percent

(a) HCl                                                        1.179                                               36.0

(b) H2SO4 1.835                                               96.0

(c) NaOH                                                    1.043                                                 4.00

           (d) Na2CO3 1.040 4.00

Explanation / Answer

a)
Percentage = 36% means 36 g of HCl in 100g of solution
volume of the solution = 100/1.179 = 84.8176 mL
normality = (36/(36.5/1))*(1000/84.8176) = 11.628 N
b)
Percentage = 96% means 96 g of H2SO4 in 100g of solution
volume of the solution = 100/1.835 = 54.496 mL
normality = (96/(98/2))*(1000/54.496) = 35.95 N
c)
Percentage = 4% means 4 g of NaOH in 100g of solution
volume of the solution = 100/1.043 = 95.877 mL
normality = (4/(40/1))*(1000/95.877) = 1.043 N
d)
Percentage = 4% means 4 g of Na2CO3 in 100g of solution
volume of the solution = 100/1.04 = 96.154 mL
normality = (4/(108/2))*(1000/96.154) = 0.7704 N

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