The molality of 1M `NaNO_(3)` solution is (d=1.25 g/ml)

Options

A. 0.8 m
B. 0.858 m
C. 1.6 m
D. 1 M `Na_(2)SO_(4)`

Answer

To find the molality of a 1M NaNO₃ solution with a density of 1.25 g/mL, we will follow these steps:

### Step 1: Determine the mass of NaNO₃ in the solution.
Since the solution is 1M, it means there is 1 mole of NaNO₃ in 1 liter (1000 mL) of solution.

**Molar mass of NaNO₃:**
- Sodium (Na): 23 g/mol
- Nitrogen (N): 14 g/mol
- Oxygen (O): 16 g/mol (3 O atoms)

Calculating the molar mass:
\[ \text{Molar mass of NaNO₃} = 23 + 14 + (3 \times 16) = 23 + 14 + 48 = 85 \text{ g/mol} \]

Thus, the mass of NaNO₃ in 1 liter of solution is:
\[ \text{Mass of NaNO₃} = 1 \text{ mole} \times 85 \text{ g/mol} = 85 \text{ g} \]

### Step 2: Calculate the mass of the solution.
Using the density of the solution (1.25 g/mL), we can find the mass of 1000 mL of solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \text{ g/mL} \times 1000 \text{ mL} = 1250 \text{ g} \]

### Step 3: Calculate the mass of the solvent (water).
The mass of the solvent can be found by subtracting the mass of the solute (NaNO₃) from the mass of the solution:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of NaNO₃} = 1250 \text{ g} - 85 \text{ g} = 1165 \text{ g} \]

### Step 4: Convert the mass of the solvent to kilograms.
Since molality is defined as moles of solute per kilogram of solvent, we convert the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent in kg} = \frac{1165 \text{ g}}{1000} = 1.165 \text{ kg} \]

### Step 5: Calculate the molality of the solution.
Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]
We already know that there is 1 mole of NaNO₃, so:
\[ m = \frac{1 \text{ mole}}{1.165 \text{ kg}} \approx 0.858 \text{ mol/kg} \]

### Final Answer:
The molality of the 1M NaNO₃ solution is approximately **0.858 mol/kg**.

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