Calculate the amount of sodium chloride (in g) which must be added to 1000 mL of water so that its freezing point is depressed by 0.744K. For water, `K_(f)` = 1.86 K/m. Assume density of water to be 1 g `mL^(-1)`.

Explanation

To calculate the amount of sodium chloride (NaCl) needed to depress the freezing point of 1000 mL of water by 0.744 K, we will use the formula for freezing point depression:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) = depression in freezing point (0.744 K)
- \(i\) = van 't Hoff factor (for NaCl, \(i = 2\) because it dissociates into Na\(^+\) and Cl\(^-\))
- \(K_f\) = cryoscopic constant for water (1.86 K/m)
- \(m\) = molality of the solution

### Step 1: Calculate the molality (m)

First, we need to rearrange the formula to solve for molality (m):

\[ m = \frac{\Delta T_f}{i \cdot K_f} \]

Substituting the values:

\[ m = \frac{0.744}{2 \cdot 1.86} \]

Calculating the denominator:

\[ 2 \cdot 1.86 = 3.72 \]

Now, substituting back into the equation:

\[ m = \frac{0.744}{3.72} \approx 0.1994 \, \text{mol/kg} \]

### Step 2: Calculate the mass of water in kg

Since we have 1000 mL of water and the density of water is 1 g/mL, the mass of water is:

\[ \text{mass of water} = 1000 \, \text{mL} \times 1 \, \text{g/mL} = 1000 \, \text{g} = 1 \, \text{kg} \]

### Step 3: Calculate the number of moles of NaCl needed

Using the definition of molality, we know:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

Rearranging gives us:

\[ \text{moles of NaCl} = m \cdot \text{mass of solvent in kg} \]

Substituting the values:

\[ \text{moles of NaCl} = 0.1994 \, \text{mol/kg} \cdot 1 \, \text{kg} = 0.1994 \, \text{mol} \]

### Step 4: Calculate the mass of NaCl required

Now, we need to find the mass of NaCl using its molar mass. 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 we can calculate the mass of NaCl:

\[ \text{mass of NaCl} = \text{moles of NaCl} \cdot \text{molar mass of NaCl} \]

Substituting the values:

\[ \text{mass of NaCl} = 0.1994 \, \text{mol} \cdot 58.5 \, \text{g/mol} \approx 11.7 \, \text{g} \]

### Final Answer

The amount of sodium chloride (NaCl) that must be added to 1000 mL of water to depress its freezing point by 0.744 K is approximately **11.7 grams**.

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