How many grams of glucose, C 6H 12O 6, (molar mass 180.2 g/mol) must be added to 656 g water to raise the boiling point of water by 1.68 °C? K b of water is 0.512 °C/m.

We must add 387.9 grams of glucose

Explanation:

Step 1: Data given

Molar mass glucose = 180.2 g/mol

Mass of water = 656 grams

Boiling point will raise 1.68 °C

Kb of water = 0.512 °C/m

Step 2: Calculate molality

ΔT = i*kb * m

⇒ ΔT = The boiling point elevation = 1.68 °C

⇒ i = the van't Hoff factor = 1

⇒ Kb = The boiling point elevation constant = 0.512 °C/m

⇒ m = molality = moles glucose / mass water

m = 1.68 / 0.512 °C/m

m = 3.28125 molal

Step 3: Calculate moles glucose

Molality = moles glucose / mass water

3.28125 molal = moles glucose / 0.656 kg

Moles glucose = 2.1525 moles

Step 4: Calculate mass glucose

Mass glucose = moles glucose * molar mass glucose

Mass glucose = 2.1525 moles * 180.2 g/mol

Mass glucose = 387.9 grams

We must add 387.9 grams of glucose

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.