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 103
(3) 2.5 x 103 (4) 1.25 x 10-2​

Answer 1

Answer:

The correct answer is option 4.

Explanation:

100 ml of 0.1 M $CaCl_2$ solution:

Moles of calcium chloride : n

Volume of the solution = 100 mL = 0.1 L

$Molarity=\frac{n}{V(l)}$

$n=0.1 M\times 0.1 L=0.01 mol$

25 ml of 0.3 N $AgNO_3$ solution

$Normality=\text{n-factor}\times Molarity$

n-factor = Total charge on cation or anion

Here n-factor = 1 (+1 charge on silver ion and -1 ion nitrate ion)

$0.3N=1\times Molarity$

Molarity = 0.3 M

Moles of $AgNO_3$ = n'

Volume of $AgNO_3$ solution = 0.25 mL = 0.025 L

$n'= 0.3M\times 0.025 L=0.0075 mol$

$CaCl_2+2AgNO_3\rightarrow 2AgCl+Ca(NO_3)_2$

According to reaction , 2 moles of silver nitrate reacts with 1 mol of calcium chloride .

Then  1 mol of calcium chloride reacts withes of silver nitrate will react with:

$\frac{1}{2}\times 0.0075 mol=0.00375 mol$ calcium chloride

Moles of calcium chloride left unreacted =

0.01 mol - 0.00375 mol = 0.00625 mol

1 mole of calcium chloride has 2 moles of chloride ions.So, 0.00625 moles of calcium chloride, we have:

$2\times 0.00625 mol = 0.0125 = 1.25\times 10^{-2} moles$