Question

a) Explain how to prepare 25 liters of a 0.10 M BaCl2 solution, starting with solid BaCl2.

b) Specify the volume of the solution in (a) needed to get 0.020 mol of BaCl2.​

Answer 1

Answer:

Concentration of KCl = 80g/L

⇒ 80 g of KCl is present in 1L of solution.

Concentration of KCl in MOLARITY is M₁ = 80/74.5 = 1.0738M

KCl → K+ + ‎Cl- ⇒ [KCl] = [Cl-]

Concentration of Cl- in MOLARITY is M₁ = 80/74.5 = 1.0738M

BaCl2 → Ba2+ + 2Cl-

Let concentration of BaCl2 be x molar

⇒ [Cl-] = 2times [BaCl2] as two moles of Cl- is produced from 1 mole of BaCl2

It is given concentration of Cl- does not change.

[Cl-] = 1.0738M ⇒ [BaCl2] = 1.0738/2 = 0.5369M

By molarity equation ,

MOLARITY = Given mass/(molar mass * volume (L))

Given mass = 0.5369*208.5*250/1000 = 27.98 ==== 27.92g ( molar mass = 208.5g,

volume = 250/1000 L)

Explanation: