# Calculate the molarity of a solution of CaCl2 if on chemical analysis it is found that 500 mL of CaCl2 solution contain 1.505 × 10^23 Cl- ions

Calculate the molarity of a solution of CaCl2 if on chemical analysis it is found that 500 mL of CaCl2 solution contain 1.505 × 1023 Cl ions

Given Values

Vsol = 500 ml

Cl–  ion in 500 ml = 1.505 × 1023 ions

CaCl2    ⇌   Ca+2   +  2Cl–
1mole      1mole    2mole

6.02 × 1023 Cl ions = 1 mole

1.505 × 1023 Cl ions = $$\frac{1.505 \times 10^{23}}{6.02 \times 10^{23}}$$ = 0.25 moles

$$n_{CaCl_{2}} = n_{Ca^{+2}} = \frac{n_{Cl^{-}}}{2}$$ $$=\frac{0.25}{2}$$ = 0.125 mole

Calculate

Molarity = $$\frac{n_{B} \times 1000}{V_{sol}}$$

Molarity = $$\frac{0.125 \times 1000}{500}$$

Molarity = 0.25 mol/L