What mass of MgCl₂ in grams must be added to 250.0 mL of a 0.25 M MgCl₂ solution to produce a 0.40 M solution, assuming no change of volume upon addition?

Question

Quizplus · Accepted Answer

First, we can use the formula for dilution to determine how much MgCl2 is needed to make the new solution:

M1V1 = M2V2

Where M1 = 0.25 M, V1 = 250.0 mL, M2 = 0.40 M, and V2 = the total volume of the new solution (which we assume is the same as the initial volume, 250.0 mL):

(0.25 M)(250.0 mL) = (0.40 M)(250.0 mL + x)

Solving for x, we get:

x = 90.0 mL

So we need to add 90.0 mL of MgCl2 to the initial solution.

To convert mL to grams, we need to use the density of the solution. The density of a 0.25 M MgCl2 solution is approximately 1.06 g/mL, so 90.0 mL of the solution has a mass of:

90.0 mL x 1.06 g/mL = 95.4 g

However, we only need to add MgCl2, not the entire solution. The molar mass of MgCl2 is 95.21 g/mol, so 95.4 g of the solution contains:

95.4 g / 95.21 g/mol = 1.001 mol MgCl2

Finally, we can use the molarity of the initial solution to determine the number of moles of MgCl2 in the 250.0 mL of solution:

0.25 M x 0.2500 L = 0.0625 mol MgCl2

To make a 0.40 M solution, we need a total of:

(0.40 M)(0.2500 L) = 0.1000 mol MgCl2

So we need to add:

0.1000 mol MgCl2 - 0.0625 mol MgCl2 = 0.0375 mol MgCl2

of MgCl2. Since we know that 1.001 mol of MgCl2 is present in 95.4 g of the solution, we can find the mass of MgCl2 needed:

0.0375 mol MgCl2 x (95.21 g/mol) = 3.54 g MgCl2

This corresponds to answer choice D.

What Mass of MgCl2 in Grams Must Be Added to 250.0

What Mass of MgCl₂ in Grams Must Be Added to 250.0