Question

100 ml of 0.1 M CaCl2 solution is mixed with
25 ml of 0.3 N AgNO3 solution. Moles of
chloride ions remain unreacted is
(1) 2.5
(2) 5 x 10-3
(3) 2.5 x 10-3
(4) 1.25 x 10-2​

Answer 1

Answer:

Explanation:refer the images

Answer 2

answer : option (2) 5 × 10^-3

explanation : mole of CaCl2 = volume of CaCl2 × molarity of CaCl2

= (100/1000)L × 0.1M

= 0.01 mol

mole of AgNO3 = volume of AgNO3 × molarity of AgNO3

= (25/1000)L × 0.3 M

= 7.5/1000

= 0.0075 mol

[ as normality = n × molarity , 0.3 N = 1 × molarity ⇒molarity = 0.3 M]

CaCl2 + 2AgNO3 ⇔2AgCl + Ca(NO3)2

here it is clear that, one mole of CaCl2 reacts with 2 moles of AgNO3.

so, 0.01 mol of CaCl2 reacts with 2 × 0.01 = 0.02 mol of AgNO3.

but given, no of moles of AgNO3 = 0.0075 mol

so, AgNO3 is limiting reagent.

so no of CaCl2 remaining = 0.01 - 0.0075

= 0.0025 mol

so, no of moles of Cl– ions remain interacted = 2 × 0.0025 = 0.0050mol

= 5 × 10^-3 mol

hence, option (2) is correct choice.