It required 0.7g of hydrocarbon (A) to react completely with 2.0g of `Br_(2)`. On treatment of 'A' with HBr it gave monobromo alkane (B). The same compound (B) was obtained when (A) was treated with HBr in presence of peroxide. Calculate the molecular weight of 'B'.

Answer

To solve the problem step by step, we will analyze the information given and perform the necessary calculations to find the molecular weight of compound B.

### Step 1: Calculate the number of moles of Br2
We know that the mass of Br2 used is 2.0 g. The molar mass of Br2 (bromine) can be calculated as follows:
- Molar mass of Br = 80 g/mol
- Therefore, Molar mass of Br2 = 2 × 80 g/mol = 160 g/mol

Now, we can calculate the number of moles of Br2:
\[ \text{Number of moles of Br2} = \frac{\text{mass of Br2}}{\text{molar mass of Br2}} = \frac{2.0 \text{ g}}{160 \text{ g/mol}} = 0.0125 \text{ moles} \]

### Step 2: Determine the number of moles of hydrocarbon A
From the reaction, we know that one mole of hydrocarbon A reacts with one mole of Br2. Therefore, the number of moles of hydrocarbon A is also:
\[ \text{Number of moles of A} = 0.0125 \text{ moles} \]

### Step 3: Calculate the molecular weight of hydrocarbon A
We can find the molecular weight of hydrocarbon A using the formula:
\[ \text{Molecular weight of A} = \frac{\text{mass of A}}{\text{number of moles of A}} = \frac{0.7 \text{ g}}{0.0125 \text{ moles}} = 56 \text{ g/mol} \]

### Step 4: Determine the molecular formula of hydrocarbon A
To find the number of carbon atoms in hydrocarbon A, we divide its molecular weight by the molar mass of carbon (12 g/mol):
\[ \text{Number of carbon atoms} = \frac{56 \text{ g/mol}}{12 \text{ g/mol}} \approx 4.67 \]
We round this to 4 carbon atoms.

Next, we calculate the number of hydrogen atoms:
\[ \text{Molecular weight of hydrogen} = 1 \text{ g/mol} \]
The total mass contributed by carbon is:
\[ 4 \text{ C} \times 12 \text{ g/mol} = 48 \text{ g/mol} \]
Thus, the mass contributed by hydrogen is:
\[ 56 \text{ g/mol} - 48 \text{ g/mol} = 8 \text{ g/mol} \]
This means there are:
\[ \text{Number of hydrogen atoms} = \frac{8 \text{ g/mol}}{1 \text{ g/mol}} = 8 \]
So, the molecular formula of hydrocarbon A is \(C_4H_8\).

### Step 5: Determine the molecular formula of compound B
When hydrocarbon A reacts with HBr, it forms a monobromoalkane B. The molecular formula of B can be derived from A:
\[ \text{Molecular formula of B} = C_4H_8Br \]
This means the molecular formula of B is \(C_4H_7Br\).

### Step 6: Calculate the molecular weight of compound B
Now, we can calculate the molecular weight of compound B:
\[ \text{Molecular weight of B} = (4 \times 12) + (7 \times 1) + (1 \times 80) = 48 + 7 + 80 = 135 \text{ g/mol} \]

### Final Answer
The molecular weight of compound B is **135 g/mol**.

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