The solubility product `(K_(sp))` of the salts of types `MX,MX_(2)andM_(3)X` at temperature 'T' are `4.0xx10^(-8),3.2xx10^(-14)and2.7xx10^(-15)` respectively. Solubilities (mol `dm^(-3)`) of the salts at temperature 'T' are in the order

Options

A. `MXgtMX_(2)gtM_(3)X`
B. `M_(3)XgtMX_(2)gtMX`
C. `MX_(2)gtM_(3)XgtMX`
D. `MXgtM_(3)XgtMX_(2)`

Answer

To solve the problem, we need to find the solubility of the salts MX, MX₂, and M₃X using their solubility product constants (Ksp) and then compare their solubilities.

### Step 1: Determine the solubility of MX

For the salt MX, the dissociation can be represented as:
\[ MX \rightleftharpoons M^+ + X^- \]

The solubility product (Ksp) expression is:
\[ K_{sp} = [M^+][X^-] = s^2 \]
where \( s \) is the solubility of MX.

Given:
\[ K_{sp} = 4.0 \times 10^{-8} \]

Setting up the equation:
\[ s^2 = 4.0 \times 10^{-8} \]

Now, we solve for \( s \):
\[ s = \sqrt{4.0 \times 10^{-8}} \]
\[ s = 2.0 \times 10^{-4} \]

So, the solubility of MX (\( s_1 \)) is:
\[ s_1 = 2.0 \times 10^{-4} \, \text{mol dm}^{-3} \]

### Step 2: Determine the solubility of MX₂

For the salt MX₂, the dissociation can be represented as:
\[ MX_2 \rightleftharpoons M^{2+} + 2X^- \]

The Ksp expression is:
\[ K_{sp} = [M^{2+}][X^-]^2 = 4s^3 \]
where \( s \) is the solubility of MX₂.

Given:
\[ K_{sp} = 3.2 \times 10^{-14} \]

Setting up the equation:
\[ 4s^3 = 3.2 \times 10^{-14} \]

Now, we solve for \( s \):
\[ s^3 = \frac{3.2 \times 10^{-14}}{4} \]
\[ s^3 = 8.0 \times 10^{-15} \]

Taking the cube root:
\[ s = (8.0 \times 10^{-15})^{1/3} \]
\[ s \approx 2.0 \times 10^{-5} \]

So, the solubility of MX₂ (\( s_2 \)) is:
\[ s_2 = 2.0 \times 10^{-5} \, \text{mol dm}^{-3} \]

### Step 3: Determine the solubility of M₃X

For the salt M₃X, the dissociation can be represented as:
\[ M_3X \rightleftharpoons 3M^{3+} + X^- \]

The Ksp expression is:
\[ K_{sp} = [M^{3+}]^3[X^-] = 27s^4 \]
where \( s \) is the solubility of M₃X.

Given:
\[ K_{sp} = 2.7 \times 10^{-15} \]

Setting up the equation:
\[ 27s^4 = 2.7 \times 10^{-15} \]

Now, we solve for \( s \):
\[ s^4 = \frac{2.7 \times 10^{-15}}{27} \]
\[ s^4 = 1.0 \times 10^{-16} \]

Taking the fourth root:
\[ s = (1.0 \times 10^{-16})^{1/4} \]
\[ s = 1.0 \times 10^{-4} \]

So, the solubility of M₃X (\( s_3 \)) is:
\[ s_3 = 1.0 \times 10^{-4} \, \text{mol dm}^{-3} \]

### Step 4: Compare the solubilities

Now we have:
- \( s_1 = 2.0 \times 10^{-4} \) (for MX)
- \( s_2 = 2.0 \times 10^{-5} \) (for MX₂)
- \( s_3 = 1.0 \times 10^{-4} \) (for M₃X)

Ordering these values:
1. \( s_1 = 2.0 \times 10^{-4} \) (MX)
2. \( s_3 = 1.0 \times 10^{-4} \) (M₃X)
3. \( s_2 = 2.0 \times 10^{-5} \) (MX₂)

Thus, the order of solubility is:
\[ MX > M₃X > MX₂ \]

### Final Answer
The solubilities (mol dm⁻³) of the salts at temperature 'T' are in the order:
\[ MX > M₃X > MX₂ \]