Home
Class 12
CHEMISTRY
Solution of calcium phosphate (molecular...

Solution of calcium phosphate (molecular mass, M) in water is w g per at `25^(@)`C, its solubility product at `25^(@)`C will be approximately :-

A

`10^(9)((W)/(M))^(5)`

B

`10^(7)((W)/(M))^(5)`

C

`10^(5)((W)/(M))^(5)`

D

`10^(3)((W)/(M))^(5)`

Text Solution

Verified by Experts

The correct Answer is:
B

`{:(Ca_(3)(PO_(4))_(2),hArr,3Ca^(+2),+,2PO_(4)^(-2)),(" "S,,3S,,2S):}`
100ml have = w gm
100 ml have `(w)/(100)xx1000=(wxx10)(M)`(solubility)
Now Ksp = `[Ca^(+2)]^(3)[PO_(4)^(-2)]^(2)`
`rArr[(wxx10xx3)/(M)]^(3)[(wxx10xx2)/(M)]^(2)`
Approx `= 10^(7)((w)/(M))^(5)`
Promotional Banner

Topper's Solved these Questions

  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise INORGANIC CHEMISTRY|300 Videos
  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise ORGANIC CHEMISTRY|483 Videos
  • Chemical Equilibrium

    ALLEN|Exercise All Questions|24 Videos
  • ELECTROCHEMISTRY

    ALLEN|Exercise EXERCISE -05 [B]|38 Videos

Similar Questions

Explore conceptually related problems

Solubility of calcium phosphate (molecular mass, M ) in water is W g per 100 mL at 25^(@)C . Its solubility product at 25^(@)C will be approximately

The solubility of calcium phosphate in water is x molL^(-1) at 250^(o) C .Its solubility product is equal

The solubility of an insoluble phosphate, M_(3)(PO_(4))_(2) of molecular weight w in water is x grams per litre, its solubility product is proportional to

The ionic product of water at 25^@C is 10^(-14) its ionic product at 90^@C will be

If HCl is added to pure water at 25^(@)C the ionic product of water will be

The solubility of calomel in water at 25^(@)C is x mole // litre. Its solubility product is

If 3g of glucose (molecular mass 180) is dissolved in 60g of water at 15^(@)C , then the osmotic pressure of this solution will be :

The solubility of a sparingly soluble salt A_(x)B_(x) in water at 25^(@)C=1.4xx10^(-4)M . The solubility product is 1.1xx10^(-11) . The possibilities are