# 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

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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, class 12 chemistry question and answer, solutions

**Calculate the molarity of a solution of CaCl _{2} if on chemical analysis it is found that 500 mL of CaCl_{2} solution contain 1.505 × 10^{23} Cl^{–} ions**

**Answer:**

**Given Values**

V_{sol} = 500 ml

Cl^{– }ion in 500 ml = 1.505 × 10^{23} ions

CaCl_{2} ⇌ Ca^{+2} + 2Cl^{– }

1mole 1mole 2mole

6.02 × 10^{23} Cl^{–} ions = 1 mole

1.505 × 10^{23} 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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