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):

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the Molar Masses
Let the molar mass of element A be \( x \) (in kg/mol) and the molar mass of element B be \( y \) (in kg/mol).

### Step 2: Write the Molar Mass Expressions
1. For the compound \( AB_2 \):
- The molar mass of \( AB_2 \) = \( x + 2y \)

2. For the compound \( A_2B_2 \):
- The molar mass of \( A_2B_2 \) = \( 2x + 2y \)

### Step 3: Use the Given Information
From the problem, we know:
- The mass of 5 moles of \( AB_2 \) is \( 125 \times 10^{-3} \) kg.
- The mass of 10 moles of \( A_2B_2 \) is \( 300 \times 10^{-3} \) kg.

### Step 4: Set Up the Equations
Using the formula for moles:
\[ \text{Moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

1. For \( AB_2 \):
\[ 5 = \frac{125 \times 10^{-3}}{x + 2y} \]
Rearranging gives:
\[ x + 2y = \frac{125 \times 10^{-3}}{5} = 25 \times 10^{-3} \quad \text{(Equation 1)} \]

2. For \( A_2B_2 \):
\[ 10 = \frac{300 \times 10^{-3}}{2x + 2y} \]
Rearranging gives:
\[ 2x + 2y = \frac{300 \times 10^{-3}}{10} = 30 \times 10^{-3} \]
Simplifying gives:
\[ x + y = 15 \times 10^{-3} \quad \text{(Equation 2)} \]

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

Subtract Equation 2 from Equation 1:
\[ (x + 2y) - (x + y) = (25 \times 10^{-3}) - (15 \times 10^{-3}) \]
This simplifies to:
\[ y = 10 \times 10^{-3} \text{ kg/mol} \]

### Step 6: Find the Value of \( x \)
Substituting \( y \) back into Equation 2:
\[ x + 10 \times 10^{-3} = 15 \times 10^{-3} \]
Solving for \( x \):
\[ x = 15 \times 10^{-3} - 10 \times 10^{-3} = 5 \times 10^{-3} \text{ kg/mol} \]

### Final Answer
The molar masses of A and B are:
- Molar mass of A = \( 5 \times 10^{-3} \) kg/mol
- Molar mass of B = \( 10 \times 10^{-3} \) kg/mol