If mass of 5 mole `AB_(2)` is `125xx10^(-3)kg` and mass of 10 mole `A_(2)B_(2)` is `300xx10^(-3)kg`. Then correct molar mass of A and B respectively (in kg/mol):

Options

A. `M_(A) = 50 xx 10^(-3)` and `M_(B) = 25 xx 10^(-3)`
B. `M_(A) = 25 xx 10^(-3)` and `M_(B) = 50 xx 10^(-3)`
C. `M_(A) = 5 xx 10^(-3)` and `M_(B) = 10 xx 10^(-3)`
D. `M_(A) = 10 xx 10^(-3)` and `M_(B) = 5 xx 10^(-3)`

Explanation

To solve the problem step by step, we will first convert the given masses into a usable form, then set up equations based on the molar masses of the compounds, and finally solve for the molar masses of elements A and B.

### Step 1: Convert the mass of the compounds from milligrams to kilograms
We know that:
- 1 milligram = \(10^{-3}\) kilograms

Given:
- Mass of 5 moles of \(AB_2\) = \(125 \, \text{mg} = 125 \times 10^{-3} \, \text{kg}\)
- Mass of 10 moles of \(A_2B_2\) = \(300 \, \text{mg} = 300 \times 10^{-3} \, \text{kg}\)

### Step 2: Calculate the mass of 1 mole of each compound
For \(AB_2\):
\[ \text{Mass of 1 mole of } AB_2 = \frac{125 \times 10^{-3} \, \text{kg}}{5} = 25 \times 10^{-3} \, \text{kg} \]

For \(A_2B_2\):
\[ \text{Mass of 1 mole of } A_2B_2 = \frac{300 \times 10^{-3} \, \text{kg}}{10} = 30 \times 10^{-3} \, \text{kg} \]

### Step 3: Set up equations for the molar masses
Let:
- Molar mass of \(A\) = \(X\) kg/mol
- Molar mass of \(B\) = \(Y\) kg/mol

From the formula for \(AB_2\):
\[ X + 2Y = 25 \times 10^{-3} \quad \text{(Equation 1)} \]

From the formula for \(A_2B_2\):
\[ 2X + 2Y = 30 \times 10^{-3} \quad \text{(Equation 2)} \]

### Step 4: Simplify Equation 2
Dividing Equation 2 by 2:
\[ X + Y = 15 \times 10^{-3} \quad \text{(Equation 3)} \]

### Step 5: Solve the equations
Now we have two equations:
1. \(X + 2Y = 25 \times 10^{-3}\)
2. \(X + Y = 15 \times 10^{-3}\)

Subtract Equation 3 from Equation 1:
\[ (X + 2Y) - (X + Y) = (25 \times 10^{-3}) - (15 \times 10^{-3}) \]
This simplifies to:
\[ Y = 10 \times 10^{-3} \, \text{kg} \]

### Step 6: Substitute \(Y\) back to find \(X\)
Substituting \(Y\) into Equation 3:
\[ X + 10 \times 10^{-3} = 15 \times 10^{-3} \]
Thus,
\[ X = 15 \times 10^{-3} - 10 \times 10^{-3} = 5 \times 10^{-3} \, \text{kg} \]

### Final Results
- Molar mass of \(A\) = \(X = 5 \times 10^{-3} \, \text{kg/mol} = 5 \, \text{g/mol}\)
- Molar mass of \(B\) = \(Y = 10 \times 10^{-3} \, \text{kg/mol} = 10 \, \text{g/mol}\)

### Conclusion
The correct molar masses of A and B are:
- Molar mass of A = \(5 \, \text{kg/mol}\)
- Molar mass of B = \(10 \, \text{kg/mol}\)