Question

An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g/mol and the molar mass of water is 18.0 g/mol. A) What is the molality of the solution? B) What is the mole fraction of the solute?

Answer 1

A) 2.69 M

B) 0.059

Step-by-step explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity=
`(33.8 g solute)/(100 g solution)` x
`(1.15 g solution)/(1 ml)` x
`(1 mol solute)/(145.6 g solute)` x
`(1000 ml)/(1 L)`

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x
`(1 mol solute)/(145.6 g)` = 0.232 mol

mol H20= 66.2 g H₂O x
`(1 mol H2O)/(18 g)`

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute=
`(mol solute)/(total mol)= (mol solute)/(mol solute + mol H2O)=(0.232 mol)/(0.232 mol + 3.677 mol)`

Xsolute= 0.059