Question

500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. CaCl2 = Ca++ + 2Cl-

a) Express the concentration of the solution in % w/v

b) Express the concentration in ratio strength

c) Express the concentration in molarity (M)

d) Express the concentration in molality (m)

e) How many equivalents of calcium chloride would be in 1.5 L of the solution?

Answer 1

a) 0,2% w/v

b) r=500

c) 0,0182 M

d) 0,0145 m

e) 0,0137 equivalents

Step-by-step explanation:

a) % w/v means mass of solute in grams per 100 mililiter of solution. Thus:

%w/v=
`(1,000 g CaCl2)/(500mL)`×100 = 0,2%w/v

b) Ratio strength is a way to express concentration. For w/v is in 1g of solute r mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.

c) Molarity is moles of solute per liter of solution, thus:

1,000 g of CaCl₂ ×
`(1mol)/(110g)` = 9,09×10⁻³ moles of CaCl₂

500 mL of solution ×
`(1L)/(1000mL)` = 0,500 L of solution

M =
`(9,09x10^(-3) moles )/(0,500 L)` = 0,0182 M

d) Molality is moles of solute per kg of solution.

Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:

kg of solution = 0,500 L of solution ×
`(0,8 kg)/(1L)` = 0,625 kg of solution

m =
`(9,09x10^(-3)moles )/(0,625 kg)` = 0,0145 m

e) In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:

1,5 L ×
`(0,0182 CaCl2 moles)/(1L)` ×
`(1equivalent)/(2 moles)` = 0,0137 equivalents

I hope it helps!