Calculate the molarity of solution made by dissolving 20.6 grams of sodium sulfate, Na2SO4, in enough water to form exactly 750 mL of solution.
The correct answer and explanation is :
To calculate the molarity of a solution, we use the formula:
$$
M = \frac{n}{V}
$$
Where:
- $M$ is the molarity of the solution (in moles per liter, mol/L),
- $n$ is the number of moles of the solute,
- $V$ is the volume of the solution in liters (L).
Step 1: Calculate the molar mass of sodium sulfate (Na₂SO₄).
The formula for sodium sulfate is Na₂SO₄. To find the molar mass, we sum the atomic masses of all atoms in the formula:
- Sodium (Na): 2 atoms × 22.99 g/mol = 45.98 g/mol
- Sulfur (S): 1 atom × 32.07 g/mol = 32.07 g/mol
- Oxygen (O): 4 atoms × 16.00 g/mol = 64.00 g/mol
So, the molar mass of Na₂SO₄ is:
$$
45.98 \, \text{g/mol} + 32.07 \, \text{g/mol} + 64.00 \, \text{g/mol} = 142.05 \, \text{g/mol}
$$
Step 2: Calculate the number of moles of Na₂SO₄.
Next, we use the mass of sodium sulfate given in the problem (20.6 grams) and the molar mass we just calculated (142.05 g/mol) to find the number of moles of Na₂SO₄:
$$
n = \frac{\text{mass}}{\text{molar mass}} = \frac{20.6 \, \text{g}}{142.05 \, \text{g/mol}} = 0.145 \, \text{mol}
$$
Step 3: Convert the volume of the solution to liters.
The volume of the solution is given as 750 mL, which is 0.750 L (since 1 L = 1000 mL).
Step 4: Calculate the molarity.
Now we can substitute the values into the molarity formula:
$$
M = \frac{n}{V} = \frac{0.145 \, \text{mol}}{0.750 \, \text{L}} = 0.193 \, \text{mol/L}
$$
Final Answer:
The molarity of the solution is 0.193 mol/L.
Explanation:
Molarity is a concentration unit that expresses the amount of solute (in moles) present in one liter of solution. In this case, we first calculated the number of moles of sodium sulfate (Na₂SO₄) by dividing the given mass by the molar mass. We then divided this value by the volume of the solution in liters to obtain the molarity. The molarity tells us how many moles of Na₂SO₄ are present in each liter of the final solution.