Question

How many chloride ions are contained in a mixture of 3.00 mL of 1.50 M calcium chloride solution with 2.00 mL of 0.50 M hydrochloric acid solution?

Answer 1

no of moles of CalCl2

molarity = n of moles / volume in liters

1.5 = no of moles / 0.003L

no of moles = 1.5 x 0.003 = 0.0045 moles

one mole of CaCl2 can have one moles of Clcium and 2 moles of chloride ions

so 0.0045 moles contain 2 x 0.0045 moles of CaCl2 = 0.009 moles chlorides ions

similarly no of moles of HCl

no of moles = 0.5 x 0.002

= 0.001 moles

one mole of HCl can have one mole of Cl-

so 0.001 mole of HCl have 0.001mole of Chloride ion

now total moles = 0.001 + 0.009 = 0.01 moles

generally one mole of molecules will gave avagadro no of molecules which is 6.023 x10 ²³

0.01 moles will have 0.01 x 6.023 x 10²³

= 6.023 x 10^{21 chloride ions}