SATHEE: Equilibrium Question 187

Equilibrium Question 187

Question: The solubility of $B a S O_{4}$ in water is $2.33 \times 10^{- 3}, g m / l i t r e .$ Its solubility product will be (molecular weight of $B a S O_{4} = 233)$ [AIIMS 1998]

Options:

A) $1 \times 10^{- 5}$

B) $1 \times 10^{- 10}$

C) $1 \times 10^{- 15}$

D) $1 \times 10^{- 20}$

Show Answer

Answer:

Correct Answer: B

Solution:

The solubility of $B a S O_{4}$ in g/litre is given $2.33 \times 10^{- 3}$

$∵$ in mole/litre. $n = \frac{W}{m ., w t}$

$= 1 \times 10^{- 5}$

$=, \frac{2.33 \times 10^{- 3}}{233}$ Because $B a S O_{4}$ is a compound $K_{s p} = S^{2} = {[1 \times 10^{- 5}]}^{2}$

$= 1 \times 10^{- 10}$

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Equilibrium Question 187

Question: The solubility of BaSO4 in water is 2.33×10−3,gm/litre. Its solubility product will be (molecular weight of BaSO4=233) [AIIMS 1998]

Options:

Answer:

Solution:

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Question: The solubility of $B a S O_{4}$ in water is $2.33 \times 10^{- 3}, g m / l i t r e .$ Its solubility product will be (molecular weight of $B a S O_{4} = 233)$ [AIIMS 1998]