how much magnesium chloride MgCl2would be needed to prepare 250cm^3 of a solution of 0.1M in chloride ion?

Answer 1

Since you ask 0.1 M in Cl- the answer is :
M = moles / L
V = 250 cm^3 = 0.250 L
moles = M x L
moles Cl- = 0.1 x 0.250 = 0.0250
MgCl2 >> Mg2+ + 2 Cl-
The ratio between MgCl2 and Cl- is 1 : 2
Moles MgCl2 = 0.0250 / 2 = 0.0125
Molar mass = 95.21 g/mol
95.21 g/mol x 0.0125 mol = 1.19 g

Answer 2

Atomic weights: Mg=24 Cl=35.5 MgCl2=95

250mLMgCl2 x 0.1molMgCl2/1000mLMgCl2 x 95gMgCl2/1molMgCl2 = 2.4g MgCl2