Question

Calculate the molarity of a solution of cacl2. if on chemical analysis it is found that 200 ml of CaCl2 contains 3.01×10²² Cl- ions

Answer 1

Since CaCl2 has 2 chloride ions, the number of CaCl2 “molecules” in 200mL is 
= 3.01x10^22 ÷ 2 

= 3.01x10^22 ÷ 2 x 5 per litre 
= 7.525 x 10^22 “molecules” 

1 mole of a substance contains 6.02 x 10^23 “molecules” 

Therefore, number of “molecules” of CaCl2 / litre 

= 7.525 x 10^22 ÷ 6.02 x 10^23 
= 0.125 

CaCl2 solution is 0.125 M

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