Question

A solution was prepared by dissolving glucose (C6H12O6, MM 180) in 100.0 grams of acetic acid. The boiling point of the solution was found to be 119.81 C.The beoiling point of acetic acid is 118.10 C and the boiling point elevation constant is 3.07 C/m. How many grams of glucose were dissolved?a) 10.0gb) 15.1gc) 21.0gd) 17.8ge) 21.0g

Answer 1

a) 10.0 g

Step-by-step explanation:

The glucose is a nonvolatile compound, so when added to a solvent, it will change the boiling point of this substance, which is called ebullioscopy. The change in temperature can be calculated by:

ΔT = K*W*i

Where K is the constant, W is the molality, and i is the van't Hoff factor.

The van't Hoff factor is related to the dissociation of the compound. For molecular compounds, such as glucose, i = 1.

119.81 - 118.10 = 3.07*W

3.07W = 1.71

W = 0.5570 mol/kg

The molality can be calculated by:

W = m1/M1*m2

Where m1 is the mass of the solute in g, M1 is the molar mass of the solute, and m2 is the mass of the solvent in kg.

0.5570 = m1/180*0.1

m1 = 10.026 g ≅ 10.0 g