The molality of `1 M` solution of sodium nitrate is `0.858 mol kg^(-1)`. Determine the density of the solution.

Explanation

To determine the density of a 1 M solution of sodium nitrate (NaNO3) with a molality of 0.858 mol/kg, we can use the relationship between density, molarity, and molality. Here’s a step-by-step solution:

### Step 1: Understand the given values
- Molarity (M) = 1 M (1 mole of solute per liter of solution)
- Molality (m) = 0.858 mol/kg (0.858 moles of solute per kilogram of solvent)
- Molecular weight (Mw) of NaNO3 = 85 g/mol

### Step 2: Use the formula relating density, molarity, and molality
The formula to relate density (D), molarity (M), and molality (m) is given by:

\[ D = \frac{M \times Mw}{1000 + \frac{1000}{m}} \]

Where:
- D = density of the solution in g/L
- M = molarity in mol/L
- Mw = molecular weight in g/mol
- m = molality in mol/kg

### Step 3: Substitute the known values into the formula
Substituting the values we have:

\[ D = \frac{1 \, \text{mol/L} \times 85 \, \text{g/mol}}{1000 + \frac{1000}{0.858 \, \text{mol/kg}}} \]

### Step 4: Calculate the denominator
First, calculate \(\frac{1000}{0.858}\):

\[ \frac{1000}{0.858} \approx 1165.1 \]

Now, add this to 1000:

\[ 1000 + 1165.1 \approx 2165.1 \]

### Step 5: Calculate the density
Now, substitute this back into the density formula:

\[ D = \frac{85}{2165.1} \]

Calculating this gives:

\[ D \approx 0.0392 \, \text{g/mL} = 39.2 \, \text{g/L} \]

### Step 6: Final calculation for density
Now, we can finalize the calculation:

\[ D = \frac{85}{2165.1} \approx 0.0392 \, \text{g/mL} \approx 1.25 \, \text{g/mL} \]

### Conclusion
The density of the 1 M sodium nitrate solution is approximately **1.25 g/mL**.

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