Q119 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Chemistry
The solubility product of $\text{Ag}_2\text{CrO}_4$ is $4 \times 10^{-12}$. What is the amount of $\text{Ag}_2\text{CrO}_4$ in gram present in $100$ ml its saturated solution? (molar mass of $\text{Ag}_2\text{CrO}_4 = 331.73$ g/mol).
- A$3.3173 \times 10^{-2}$ g
- Correct answerB$3.3173 \times 10^{-3}$ g
- C$3.3173 \times 10^{-4}$ g
- D$3.3173 \times 10^{-5}$ g
Solution
Let the solubility of $\text{Ag}_2\text{CrO}_4$ be $s$ mol/L.
The dissociation reaction is:
$\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$
The solubility product $K_{sp}$ is given by:
$K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] = (2s)^2(s) = 4s^3$
Given $K_{sp} = 4 \times 10^{-12}$, we have:
$4s^3 = 4 \times 10^{-12}$
$s^3 = 10^{-12}$
$s = 10^{-4}$ mol/L
The number of moles of $\text{Ag}_2\text{CrO}_4$ in $100$ ml ($0.1$ L) of saturated solution is:
Moles $= s \times V = 10^{-4} \times 0.1 = 10^{-5}$ mol
The amount of $\text{Ag}_2\text{CrO}_4$ in grams is:
Mass $= \text{Moles} \times \text{Molar mass}$
Mass $= 10^{-5} \times 331.73 = 3.3173 \times 10^{-3}$ g
Answer: $3.3173 \times 10^{-3}$ g
The dissociation reaction is:
$\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$
The solubility product $K_{sp}$ is given by:
$K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] = (2s)^2(s) = 4s^3$
Given $K_{sp} = 4 \times 10^{-12}$, we have:
$4s^3 = 4 \times 10^{-12}$
$s^3 = 10^{-12}$
$s = 10^{-4}$ mol/L
The number of moles of $\text{Ag}_2\text{CrO}_4$ in $100$ ml ($0.1$ L) of saturated solution is:
Moles $= s \times V = 10^{-4} \times 0.1 = 10^{-5}$ mol
The amount of $\text{Ag}_2\text{CrO}_4$ in grams is:
Mass $= \text{Moles} \times \text{Molar mass}$
Mass $= 10^{-5} \times 331.73 = 3.3173 \times 10^{-3}$ g
Answer: $3.3173 \times 10^{-3}$ g
Hence, the correct answer is option: 'B'.