At 310 K, the solubility of CaF 2 in water is 2.34 × 10-3 g / 100 mL. The solubility product of CaF 2 is × 10-8( mol / L )3. (Given molar mass: .CaF 2=78 g mol -1)

Question

At 310 K, the solubility of CaF 2 in water is 2.34 × 10-3 g / 100 mL. The solubility product of CaF 2 is × 10-8( mol / L )3. (Given molar mass: .CaF 2=78 g mol -1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Solubility of CaF 2= S mole / L S =(2.34 × 10-3/0.1 × 78)=(2.34/78) × 10-2=3 × 10-4 mol / L K text sp ( CaF 2)=4 S 3=4(3 × 10-4)3 =108 × 10-12 =0.0108 × 10-8( mol / L )3

Q. At $310 K$ , the solubility of $C a F_{2}$ in water is $2.34 \times 1 0^{- 3} g /100 m L$ . The solubility product of $C a F_{2}$ is __ $\times 1 0^{- 8} (m o l / L)^{3}$ . (Given molar mass : $C a F_{2} = 78 g m o l^{- 1})$

Solubility of $C a F_{2} = S$ mole $/ L$
$S = \frac{2.34 \times 1 0 ^{- 3}}{0.1 \times 78} = \frac{2.34}{78} \times 1 0^{- 2} = 3 \times 1 0^{- 4} m o l / L$
$K_{sp} (C a F_{2}) = 4 S^{3} = 4 (3 \times 1 0^{- 4})^{3}$
$= 108 \times 1 0^{- 12}$
$= 0.0108 \times 1 0^{- 8} (m o l / L)^{3}$

Q. At 310K, the solubility of CaF2​ in water is 2.34×10−3g/100mL. The solubility product of CaF2​ is __ ×10−8(mol/L)3. (Given molar mass : CaF2​=78gmol−1)

Solubility of CaF2​=S mole /L S=0.1×782.34×10−3​=782.34​×10−2=3×10−4mol/L Ksp ​(CaF2​)=4S3=4(3×10−4)3 =108×10−12 =0.0108×10−8(mol/L)3

Q. At $310 K$ , the solubility of $C a F_{2}$ in water is $2.34 \times 1 0^{- 3} g /100 m L$ . The solubility product of $C a F_{2}$ is __ $\times 1 0^{- 8} (m o l / L)^{3}$ . (Given molar mass : $C a F_{2} = 78 g m o l^{- 1})$

Solubility of $C a F_{2} = S$ mole $/ L$
$S = \frac{2.34 \times 1 0 ^{- 3}}{0.1 \times 78} = \frac{2.34}{78} \times 1 0^{- 2} = 3 \times 1 0^{- 4} m o l / L$
$K_{sp} (C a F_{2}) = 4 S^{3} = 4 (3 \times 1 0^{- 4})^{3}$
$= 108 \times 1 0^{- 12}$
$= 0.0108 \times 1 0^{- 8} (m o l / L)^{3}$