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

Answer:

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

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