`1g` of `NaCl` is dissolved in `10g` of a solution the density of which is `1.07g//"cc"`. Find the molality and molarity of NaCl.

Explanation

To solve the problem of finding the molality and molarity of NaCl when 1g of NaCl is dissolved in 10g of a solution with a density of 1.07 g/cc, we can follow these steps:

### Step 1: Calculate the number of moles of NaCl
The molar mass of NaCl (sodium chloride) is approximately 58.5 g/mol. We can calculate the number of moles of NaCl using the formula:

\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:

\[ \text{Number of moles of NaCl} = \frac{1 \text{ g}}{58.5 \text{ g/mol}} \approx 0.0171 \text{ moles} \]

### Step 2: Calculate the weight of the solvent
The total mass of the solution is given as 10 g, and since we have added 1 g of NaCl, the mass of the solvent can be calculated as:

\[ \text{Weight of solvent} = \text{Total mass of solution} - \text{mass of NaCl} = 10 \text{ g} - 1 \text{ g} = 9 \text{ g} \]

### Step 3: Convert the weight of the solvent to kilograms
To calculate molality, we need the weight of the solvent in kilograms:

\[ \text{Weight of solvent in kg} = \frac{9 \text{ g}}{1000} = 0.009 \text{ kg} \]

### Step 4: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent:

\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Weight of solvent (kg)}} \]

Substituting the values:

\[ \text{Molality} = \frac{0.0171 \text{ moles}}{0.009 \text{ kg}} \approx 1.90 \text{ mol/kg} \]

### Step 5: Calculate the volume of the solution
To find the molarity, we need the volume of the solution. We can find the volume using the density formula:

\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]

Rearranging gives us:

\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{10 \text{ g}}{1.07 \text{ g/cc}} \approx 9.34 \text{ cc} \approx 0.00934 \text{ L} \]

### Step 6: Calculate the molarity
Molarity (M) is defined as the number of moles of solute per liter of solution:

\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution (L)}} \]

Substituting the values:

\[ \text{Molarity} = \frac{0.0171 \text{ moles}}{0.00934 \text{ L}} \approx 1.83 \text{ M} \]

### Final Results
- **Molality of NaCl**: approximately **1.90 mol/kg**
- **Molarity of NaCl**: approximately **1.83 M**