48 g of `SO_(3)` , 12.8 g of `SO_(2)` and 9.6 g of `O_(2)` are present in one litre . The respective active masses will be

Options

A. (a)1.0 , 05 and 0.3
B. (b)0.6 , 0.2 and 0.3
C. (c)0.6 , 0.4 and 0.2
D. (d)1.0 , 0.5 and 1.5

Answer

To find the respective active masses of \( SO_3 \), \( SO_2 \), and \( O_2 \) in the given problem, we will follow these steps:

### Step 1: Understand the concept of active mass
Active mass is defined as the number of moles of a substance per unit volume of the solution. The formula to calculate active mass is:
\[ \text{Active mass} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Since the volume is given as 1 liter, the active mass will simply be equal to the number of moles.

### Step 2: Calculate the number of moles for each substance
The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Given weight (g)}}{\text{Molecular weight (g/mol)}} \]

#### For \( SO_3 \):
- Given weight = 48 g
- Molecular weight of \( SO_3 \) = \( 32 + 3 \times 16 = 80 \) g/mol
\[ \text{Number of moles of } SO_3 = \frac{48}{80} = 0.6 \text{ moles} \]

#### For \( SO_2 \):
- Given weight = 12.8 g
- Molecular weight of \( SO_2 \) = \( 32 + 2 \times 16 = 64 \) g/mol
\[ \text{Number of moles of } SO_2 = \frac{12.8}{64} = 0.2 \text{ moles} \]

#### For \( O_2 \):
- Given weight = 9.6 g
- Molecular weight of \( O_2 \) = \( 32 \) g/mol
\[ \text{Number of moles of } O_2 = \frac{9.6}{32} = 0.3 \text{ moles} \]

### Step 3: Determine the active masses
Since the volume is 1 liter, the active masses are equal to the number of moles calculated above.

- Active mass of \( SO_3 = 0.6 \)
- Active mass of \( SO_2 = 0.2 \)
- Active mass of \( O_2 = 0.3 \)

### Final Result
The respective active masses of \( SO_3 \), \( SO_2 \), and \( O_2 \) are:
- \( SO_3 \): 0.6
- \( SO_2 \): 0.2
- \( O_2 \): 0.3

---