The boiling point of water of 735 torr is `99.07^@C` The mass of NaCl added in 100g water to make its boiling point `100^@C` is

Options

A. 10.68 g
B. 5.34g
C. 2.67g
D. 26.7g

Answer

To solve the problem of finding the mass of NaCl required to elevate the boiling point of water from 99.07°C to 100°C, we will follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔTb)
The elevation in boiling point (ΔTb) is given by the formula:
\[ \Delta T_b = T_{b,\text{solution}} - T_{b,\text{solvent}} \]

Where:
- \( T_{b,\text{solution}} = 100°C \)
- \( T_{b,\text{solvent}} = 99.07°C \)

Calculating ΔTb:
\[ \Delta T_b = 100°C - 99.07°C = 0.93°C \]

### Step 2: Use the boiling point elevation formula
The formula for boiling point elevation is:
\[ \Delta T_b = K_b \times m \times i \]

Where:
- \( K_b \) for water = 0.512 °C/m
- \( m \) = molality of the solution
- \( i \) = van 't Hoff factor (for NaCl, \( i = 2 \) because it dissociates into Na⁺ and Cl⁻)

### Step 3: Rearrange the formula to find molality (m)
Rearranging the formula gives:
\[ m = \frac{\Delta T_b}{K_b \times i} \]

Substituting the known values:
\[ m = \frac{0.93°C}{0.512 °C/m \times 2} \]

Calculating molality:
\[ m = \frac{0.93}{1.024} \approx 0.908 m \]

### Step 4: Calculate the number of moles of NaCl needed
Molality (m) is defined as moles of solute per kg of solvent. Since we have 100 g of water, we convert this to kg:
\[ \text{mass of solvent} = 100 g = 0.1 kg \]

Using the molality:
\[ m = \frac{\text{moles of NaCl}}{0.1 \text{ kg}} \]
\[ \text{moles of NaCl} = m \times 0.1 \text{ kg} \]
\[ \text{moles of NaCl} = 0.908 \times 0.1 = 0.0908 \text{ moles} \]

### Step 5: Calculate the mass of NaCl required
The molar mass of NaCl is:
\[ \text{Molar mass of NaCl} = 23 \text{ g/mol (Na)} + 35.5 \text{ g/mol (Cl)} = 58.5 \text{ g/mol} \]

Now, calculate the mass of NaCl:
\[ \text{mass of NaCl} = \text{moles of NaCl} \times \text{molar mass of NaCl} \]
\[ \text{mass of NaCl} = 0.0908 \text{ moles} \times 58.5 \text{ g/mol} \approx 5.31 \text{ g} \]

### Final Answer
The mass of NaCl required to elevate the boiling point of water to 100°C is approximately **5.31 grams**.