Question

What is the change in the boiling point of a solution made by dissolving 14.7 g of glucose into 150.0 ml of water?

Answer 1

there is a key piece of information that we are missing.

we need the following:

Kb of water= 0.512

the change in boiling point (ΔTb) can be calculated using the following formula:

ΔTb= Kb x m

we already have Kb, but we need to determine the molality (m).


`molality (m)= (moles glucose)/(Kg H_2O)`

1) let's convert the grams of glucose to moles using the molar mass of it. The molecule formula of glucose is C₆H₁₂O₆.

molar mass C₆H₁₂O₆= (6 x 12.0) + (12 x 1.01) + (6 x 16.0)= 180 g/mol


`14.7g ( (1 mol)/(180 grams) )= 0.0817 moles`

2) let's determine the Kilograms of water.

info:
density of water= 1.0 g/ mL or 1 grams = 1 mL
1000 grams= 1 kilogram


`150.0 mL = 150.0 grams`


`150.0grams ( (1 kg)/(1000 grams) )= 0.1500 Kg`

3) let's plug in the values to solve for molality


`molality= (0.0817 moles)/(0.1500 Kg) = 0.545 m`


finally, we can solve for change in boiling point.

ΔTb= Kb x m

ΔTb= (0.512) (0.545m)= 0.279°C