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

Solubility of CaF2 = S mol/L S = 2 . 3 4 × 1 0 − 3 0 . 1 × 7 8 = 2 . 3 4 7 8 × 1 0 − 2 = 3 × 1 0 − 4 m o l / L <!- [if gte mso 9]><xml> </xml><! [endif]-> K s p ( C a F 2 ) = 4 S 3 = 4 ( 3 × 1 0 − 4 ) 3 = 1 0 8 × 1 0 − 1 2 <!- [if gte mso 9]><xml> </xml><! [endif]-> = 0.0108 × 10-8 (mol/L)3

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Ncert Solutions Chemistry Class 12th Chemistry Class 12th

At 310 K, the solubility of CaF₂ in water is 2.34 × 10^-3 g/ 100 mL. The solubility product of CaF₂ is ____________× 10^-8 (mol/L)³. (Give molar mass : CaF₂ = 78 gmol^-1)

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alok kumar singh | Contributor-Level 10

3 months ago

Solubility of CaF₂ = S mol/L

$S = \frac{2.34 \times 1 0^{- 3}}{0.1 \times 78} = \frac{2.34}{78} \times 1 0^{- 2} = 3 \times 1 0^{- 4} m o l / L$

$K_{s p} (C a F_{2}) = 4 S^{3} = 4 {(3 \times 1 0^{- 4})}^{3} = 108 \times 1 0^{- 12}$

= 0.0108 × 10^-8 (mol/L)³

Solubility of CaF2 = S mol/L <math> <mrow> <mi>S</mi> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>4</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mo>×</mo> <mn>7</mn> <mn>8</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>4</mn> </mrow> <mrow> <mn>7</mn> <mn>8</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mn>3</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> <mi>m</mi> <mi>o</mi> <mi>l</mi> <mo>/</mo> <mi>L</mi> </mrow> </math>         <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815467611"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <msub> <mrow> <mi>K</mi> </mrow> <mrow> <mi>s</mi> <mi>p</mi> </mrow> </msub> <mrow> <mo> (</mo> <mrow> <mi>C</mi> <mi>a</mi> <msub> <mrow> <mi>F</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>4</mn> <msup> <mrow> <mi>S</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>=</mo> <mn>4</mn> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>3</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>=</mo> <mn>1</mn> <mn>0</mn> <mn>8</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>1</mn> <mn>2</mn> </mrow> </msup> </mrow> </math>       <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815467612"> </o:OLEObject></xml><! [endif]-> = 0.0108 × 10-8 (mol/L)3

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At 310 K, the solubility of CaF₂ in water is 2.34 × 10^-3 g/ 100 mL. The solubility product of CaF₂ is ____________× 10^-8 (mol/L)³. (Give molar mass : CaF₂ = 78 gmol^-1)

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