What is the molarity of a saturated solution of potassium sulfate if the solubility is 13 g per 100 g H₂O at 25 °C? The density of the solution is 1.1 g/mL A)0.14 M B)1.1 M C)0.96 M D)0.73 M E)0.79 M

Question

Quizplus · Accepted Answer

First, we can calculate the mass of potassium sulfate that can dissolve in 100 g of the solvent (water) at 25°C:

13 g K2SO4 / 100 g H2O = 0.13 g K2SO4 / g H2O

Next, we need to consider the density of the solution. We know that the density is 1.1 g/mL, which means that 1 mL of the solution has a mass of 1.1 g. This can be rearranged to give us the volume of 100 g of the solution:

100 g / 1.1 g/mL = 90.9 mL

Now we can use the molarity formula, which is:

Molarity = moles of solute / volume of solution in L

We know the mass of solute (K2SO4) that can dissolve in 100 g of water, so we can convert that to moles:

0.13 g K2SO4 x 1 mol K2SO4 / 174.26 g K2SO4 = 0.000747 mol K2SO4

We also know the volume of the solution that contains that amount of solute, so we can convert mL to L:

90.9 mL x 1 L / 1000 mL = 0.0909 L

Now we can substitute these values into the molarity formula:

Molarity = 0.000747 mol / 0.0909 L ≈ 0.00773 M

However, this is the molarity of a solution that has 100 g of water. We actually have a solution that has a density of 1.1 g/mL, which means that the volume we calculated (90.9 mL) includes both the water and the potassium sulfate. We need to adjust for this by dividing by the density:

90.9 mL / 1.1 g/mL ≈ 82.6 g of solution

So we actually have 82.6 g of solution that contains 0.000747 mol of K2SO4. Now we can calculate the molarity:

Molarity = 0.000747 mol / (82.6 g / 174.26 g/mol) ≈ 0.73 M

Therefore, the correct answer is D) 0.73 M.

What Is the Molarity of a Saturated Solution of Potassium