Which solution contains the largest number of moles of chloride ions?

a. 30.00 ml of 0.100m cacl2

b. 10.0 ml of 0.500m bacl2

c. 25.00 ml of 0.400m kcl

d. 4.00 ml of 1.000m nacl

e. 7.50 ml of 0.500m fecl3?

Molarity = number of moles of solute/liters of solution

number of moles of solute = molarity x liters of solution

Part (a) : 30.00 ml of 0.100m Cacl2

number of moles of CaCl2 = 0.1 x 0.03 = 3x10^-3 moles

1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine

Part (b) : 10.0 ml of 0.500m bacl2

number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles

1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine

Part (c) : 4.00 ml of 1.000m nacl

number of moles of NaCl = 1 x 0.004 = 0.004 moles

1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine

Part (d) : 7.50 ml of 0.500m fecl3

number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles

1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine

Based on the above calculations, the correct answer is (d)