Let's go through the steps to find the mass percent and molarity of a solution containing 8.000 grams of Mg(NO3)2 (magnesium nitrate) in 200.00 milliliters of water.
### Step 1: Calculate the mass of the solution
Given:
- Mass of solute (Mg(NO3)2) = 8.000 g
- Volume of solution = 200.00 mL
- Density of water = 1.00 g/mL
First, we convert the volume of the solution to the mass of the solution. Since the density of water is 1.00 g/mL, the mass of 200.00 mL of water is:
[tex]\[ \text{Mass of solution} = \text{Volume of solution} \times \text{Density of water} \][/tex]
[tex]\[ \text{Mass of solution} = 200.00 \text{ mL} \times 1.00 \text{ g/mL} = 200.00 \text{ g} \][/tex]
### Step 2: Calculate the mass percent of the solute in the solution
The mass percent is calculated using the formula:
[tex]\[ \text{Mass percent} = \left(\frac{\text{Mass of solute}}{\text{Mass of solution}}\right) \times 100 \][/tex]
Substitute the values:
[tex]\[ \text{Mass percent} = \left(\frac{8.000 \text{ g}}{200.00 \text{ g}}\right) \times 100 = 4.0\% \][/tex]
### Step 3: Convert the volume of solution to liters
Given:
- Volume of solution = 200.00 mL
Convert milliliters to liters:
[tex]\[ \text{Volume of solution in liters} = \frac{\text{Volume of solution in milliliters}}{1000} \][/tex]
[tex]\[ \text{Volume of solution in liters} = \frac{200.00 \text{ mL}}{1000} = 0.2 \text{ L} \][/tex]
### Step 4: Calculate the moles of solute (Mg(NO3)2)
Given:
- Mass of solute (Mg(NO3)2) = 8.000 g
- Molar mass (MW) of Mg(NO3)2 = 148.31 g/mol
Moles of solute is calculated using the formula:
[tex]\[ \text{Moles of solute} = \frac{\text{Mass of solute}}{\text{Molar mass of solute}} \][/tex]
Substitute the values:
[tex]\[ \text{Moles of solute} = \frac{8.000 \text{ g}}{148.31 \text{ g/mol}} \approx 0.05394 \text{ mol} \][/tex]
### Step 5: Calculate the molarity of the solution
Molarity (M) is calculated using the formula:
[tex]\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \][/tex]
Substitute the values:
[tex]\[ \text{Molarity} = \frac{0.05394 \text{ mol}}{0.2 \text{ L}} \approx 0.2697 \text{ mol/L} \][/tex]
### Summary
- Mass Percent: The mass percent of Mg(NO3)2 in the solution is 4.0%.
- Molarity: The molarity of the Mg(NO3)2 solution is approximately 0.2697 mol/L.
