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Olenka [21]
3 years ago
15

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?
Chemistry
1 answer:
vodomira [7]3 years ago
4 0

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

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

molarity = mol solute/ 1 L solution

molarity= \frac{33.8 g solute}{100 g solution} x \frac{1.15 g solution}{1 ml} x \frac{1 mol solute}{145.6 g solute} x \frac{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 \frac{1 mol solute}{145.6 g} = 0.232 mol

mol H20= 66.2 g H₂O x \frac{1 mol H2O}{18 g}

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

Xsolute=\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}

Xsolute= 0.059

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