Question

You prepare a solution “X” by adding the following ingredients to 1L of H2O: 125 mL of solution “A” (1.6 M compound “A”), 50 mL of solution “B” (20 % (m/v) compound “B”), and 25 mL of solution “C” (10g/L of compound “C”.) The properties of each ingredient are as follows: Compound A: MW 200g/mole, density 1.2g/mL, density of 1.6 M soln. 1.05g/mL Compound B: MW 150g/mole, density 1.3g/mL, density of 20% soln. 1.1g/mL Compound C: MW 35g/mole, density 1.15g/mL, density of 10g/L soln. 1.03g/mL Density of solution X: 1.25g/mL

1. What is the final molarity of compound “B” in solution X?

Answer 1

Solution X contain 50 ml of 20% (m/v) solution of compound B.

Now, we have to find out how many grams of compound B is present in 50 ml of solution,

We can find it out by doing, 50 * (20/100) = 10 g.

Now, we know that the molecular weight (MW) of compound B is 150 g/mole (given).

So, now we can find out 10 g of compound B corresponds to how many moles by dividing it by the MW.

10 g of compound B = 10/150 = 1/15 mole.

Now, lets take a look on what will be the total volume of solution X after adding water, solution of A, B and C.

Total volume of solution X = (1 + 0.125+ 0.05+ 0.025) L = 1.2 L.

Now to find out what is the molarity of the compound B in the solution X, you have to find out how many moles of compound B is present per liter of X solution. (According to the definition of Molarity).

We know that 10 g or 1/15 moles of compound B is present in 1.2 L of X solution.

So, in 1 L of solution X, (1/15)/1.2 = 1/18 moles of compound B is present.

So, the molarity of compound B in solution X is 1/18.