Question

how much gram of bacl2.8h2o is required to prepare 0.1 m in 200 ml solution

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)

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Answer 2

Answer:

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