Question

2. How many grams of Ba(OH)2.8H20 are needed for 250 ml of solution in which (OH) = 0.18 M
(molecular weight of Ba(OH)2.8H2O = 315.4)
(A) 28.7
(C) 7.1
(B) 14.2
(D) 3.8

Answer 1

we have to find the weight of Ba(OH)₂.8H₂O are needed for 250 ml of solution in which [ (OH)¯ ] = 0.18 M. (molecular weight of Ba(OH)₂.8H₂O = 315.4 g/mol).

solution : given, molarity of [(OH)¯] = 0.18 M

volume of solution = 250 ml

no of moles [(OH)¯] = molarity × volume of L

= 0.18 M × (250 /1000) L

= 0.045 mol

as it is clear that 2 moles of [ (OH)¯] in one mole of Ba(OH)₂.8H₂O .

so, 0.045 mol of [(OH)¯] in 0.045/2 = 0.0225 mol of Ba(OH)₂.8H₂O.

weight of Ba(OH)₂.8H₂O = 0.0225 × 315.4

= 7.0965 ≈ 7.1 g

Therefore the weight of Ba(OH)₂.8H₂O is 7.1 g