Question

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) What is the mass percent of sucrose in this solution?c) What is the mole fraction of sucrose in this solution?

Answer 1

a) 1,74 molal

b) 37,2 %

c) 0,03

Step-by-step explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

`Molarity = M = (1,22 mol solute)/(lts solution)`

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

`1 lts solution * (1,22 moles solute)/(lts solution)=<strong>1,22 moles solute`

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

`1 lts solution * (1,12 grs solution)/(mL solution)*(1000 mL solution)/(1 lts solution) = 1120 grs of solution`

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

`1,22 mol solute*(342 grs solute)/(1 mol solute) = 417,24 grs solute`

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

`molality = m = (1,22 mol solute)/(0,70276 kgs solvent)= <strong>1,74 molal`

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

`%(w/w) = (417,24 grs solute)/(1120 grs solution) = <strong>37,2%`

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

`702,76 grs solvent*(1 mol solvent)/(18 grs solvent) = 39,04 moles solvent`

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

`Mol frac solute = (1,22 mol solute)/(40,26 mol solution)= 0,03`