Question

1- Calculate the amount of solute, solvent, and/or solution needed to produce the following aqueous solutions:

a) 50.0 ml of a 0.050 M aqueous acetic acid solution, CH3COOH

b) 350.0 grams of a 25% w/w aqueous solution of calcium chloride, CaCl2

c) 100.0 mL of a 0.25 M nitric acid, HNO3 aqueous solution from a 0.75 M

Answer 1

a) 50.0 ml of a 0.050 M aqueous acetic acid solution, CH₃COOH

Molar mass of acetic acid = 60.052 g/mol

Mass of acetic acid required = Molarity x Volume x Molar mass

                                              = 0.05 M x 0.05 L x 60.052 g/mol = 0.15 g

0.15 g of acetic acid dissolved in 50 mL of water will produce 0.05 M acetic acid solution

b) 350.0 grams of a 25% w/w aqueous solution of calcium chloride, CaCl₂

25g CaCl₂ in 100 g H₂O = 25% w/w

Let the mass of CaCl2 = x

Mass of H2O = y

Therefore,

(x/y) = 25 % = 25/100

(x/y) = 25/100 ---------------(1)

Required,

x + y = 350 g

Therefore, x = 350 -y

Substituting the value of x in equation (1)

(350-y)/y = 25/100

35000 -100y-25y =0

35000-125 y =0

y = Mass of water required = 280 g

Mass of CaCl2 required = 350 g- 280 g = 70 g

To prepare 350.0 grams of a 25% w/w aqueous solution of calcium chloride, mix 70 g of CaCl₂ with 280 g of H₂O

c) 100.0 mL of a 0.25 M nitric acid, HNO₃ aqueous solution from a 0.75 M

M₁V₁ = M₂V₂

M₁= 0.25 M

V₁ = 100 mL

M₂ = 0.75 M

V₂ = Volume of 0.75 M HNO₃ required = (0.25 x 100)/0.75 = 33.33 mL

33.33 mL of 0.75 M HNO₃ diluted to 100 mL will produce 100 mL of 0.25 M nitric acid