`CHCl_3` and HF lead to the formation of a compound of fluorine of molecular weight 70. The compound is

Options

A. Fluoroform
B. Fluorine monoxide
C. Fluorine dioxide
D. Fluoromethanol

Explanation

To solve the problem of identifying the compound formed from the reaction of CHCl₃ (chloroform) and HF (hydrofluoric acid) that has a molecular weight of 70, we can follow these steps:

### Step 1: Analyze the reactants
- **CHCl₃ (Chloroform)** has the molecular formula CCl₃H and a molecular weight of approximately 119.38 g/mol.
- **HF (Hydrofluoric acid)** has the molecular formula HF and a molecular weight of approximately 20.01 g/mol.

### Step 2: Determine the possible reaction
When chloroform reacts with hydrofluoric acid, we can expect a substitution reaction where chlorine atoms in chloroform are replaced by fluorine atoms from HF.

### Step 3: Calculate the molecular weight of the product
- The molecular weight of the product is given as 70 g/mol.
- To find the possible product, we can assume that one of the chlorine atoms in CHCl₃ is replaced by a fluorine atom from HF.

### Step 4: Formulate the product
- If we replace one chlorine atom in CHCl₃ with a fluorine atom, the resulting compound would be:

\[ \text{CCl}_2\text{F}_1\text{H} \]

- The molecular formula of this compound is CCl₂HF.

### Step 5: Calculate the molecular weight of the product
- The molecular weight of CCl₂HF can be calculated as follows:
- Carbon (C): 12.01 g/mol
- Chlorine (Cl): 2 × 35.45 g/mol = 70.90 g/mol
- Hydrogen (H): 1.01 g/mol
- Fluorine (F): 19.00 g/mol

Adding these together:
\[ 12.01 + 70.90 + 1.01 + 19.00 = 102.92 \text{ g/mol} \]
This does not match the required molecular weight of 70 g/mol.

### Step 6: Consider another substitution
If we replace two chlorine atoms with fluorine atoms, we get:
\[ \text{CF}_2\text{H} \]
- The molecular formula of this compound is CF₂H.

### Step 7: Calculate the molecular weight of CF₂H
- The molecular weight of CF₂H can be calculated as follows:
- Carbon (C): 12.01 g/mol
- Fluorine (F): 2 × 19.00 g/mol = 38.00 g/mol
- Hydrogen (H): 1.01 g/mol

Adding these together:
\[ 12.01 + 38.00 + 1.01 = 51.02 \text{ g/mol} \]
This still does not match the required molecular weight.

### Step 8: Identify the correct compound
After considering various combinations, the compound that fits the molecular weight of 70 g/mol is:
\[ \text{CClF}_2 \]
- The molecular weight of CClF₂ can be calculated as follows:
- Carbon (C): 12.01 g/mol
- Chlorine (Cl): 35.45 g/mol
- Fluorine (F): 2 × 19.00 g/mol = 38.00 g/mol

Adding these together:
\[ 12.01 + 35.45 + 38.00 = 85.46 \text{ g/mol} \]
This also does not match.

### Conclusion
The compound that matches the molecular weight of 70 g/mol is likely to be **C₂H₃F** (Vinyl fluoride), which can be formed through the reaction of CHCl₃ and HF under specific conditions.

### Final Answer
The compound formed is **C₂H₃F** (Vinyl fluoride).