Question

a solution was prepared dissolving glucose in 100g of acetic acid the solutions boiling point is 119.81C the boiling point of pure acetic acid is 118.10C and the molal elevation constant is 3.07c/molal how many grams of glucose were used?

Answer 1

Boiling point elevation = Boiling point of solution-boiling point of solvent= 119.81-118.10=1.71 deg.c

we know that boiling point elevation= kb*m

kb= molal elevation constant and m= molality

1.71= 3.07*m

m= 0.56

molality = moles of solute. kg of solvent

solvent =acetic acid and its mass is 100g= 0,1 kg

1 kg contains 0.56 moles of solute ( glucose)

0.1 kg contains 0.56*0.1=0.056 moles of solute (Glucose is the solute)

moles= mass/Molecular weight

Mass= moles* Molecular weight

Mass of solutee =0.056*180=10.08gms