Question

How much BaCl2 would be needed to make 250 mL of a solution having same concentration of Cl– as the one containing 3.78 g of NaCl per 100 mL

Answer 1

First of we have to take NaCl.
NaCl ⇄ Na⁺ + Cl⁻
∴ [NaCl ] = [Cl⁻] = [Na⁺]
Now, molarity of NaCl = mole of NaCl /volume of solution in L
mole of NaCl = 3.78/58.5 = 0.0646
volume = 100ml = 0.1L
∴ molarity = 0.0646/0.1 = 0.646
∴ concentration of Cl⁻ = [Cl⁻] = 0.646

Again, BaCl₂ ⇄ Ba⁺ + 2Cl⁻
∴ [BaCl] = [Ba⁺] = [Cl⁻]/2
From above concentration of Cl⁻= 0.646 M
∴ [BaCl₂] = [Cl⁻]/2 = 0.646/2 = 0.323M

Now, molarity of BaCl₂ = weight of BaCl₂/molar weight of BaCl₂ × volume in L
0.323 = weight of BaCl₂/(208.3 × 0.25) [ ∵ molar weight of BaCl2 = 208.3g/mol and volume of solution = 250ml = 0.25L
∴0.323 × 208.3 × 0.25 = weight of BaCl2
weight of BaCl2 = 16.8g

Hence, weight of BaCl₂ = 16.8g

Answer 2

Answer:

weight of BaCl2 is 16.8g

Explanation: