3a 291 g of CrNO339H20 FW 400 gmol was dissolved in water a

3a) 2.91 g of Cr(NO3)3·9H20 (FW = 400 g/mol) was dissolved in water and diluted to 100 mL in a volumetric flask. a) What is the molarity of solution? (4 pts) b) What is the Molarity of Cr(III)? (4 pts) c) Calculate the equivalent weight of Cr(NO3)3·9H2O. (4 pts) d) Calculate the normality of the solution. (4 pts) e) What is the concentration of solution as percent by mass / volume (m/v)? (4 pts) f) What is the concentration of Cr(III as part per million (ppm)? (4 pts)

Solution

a) Molarity = moles of solute/volume of solution(L)

Moles of solute = given mass / molar mass= 2.91g/400g/mol

=0.00728 mol

So, Molarity =0.00728 mol/0.1L=0.0728 M

b) Since the 1 molecule of compound has only 1 Cr atom

So, concentration of solution = concentration of Cr ion

That is , Molarity of Cr(III) = 0.0728 M

c) Equivalent mass=molecular mass÷(acidity /basicity/number of electrons gained or lost )

So, equivalent mass = 400/3(because Credit can loose 3 electrons)

Equivalent mass = 133.33

d) Normality = number of gram equivalent /volume of solution in litre

Number of gram equivalent =given mass/equivalent mass

=2.91/133.33=0.0218g-eqiv

So,normality = 0.0218g-equiv/0.1L= 0.218 g-equiv/L