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

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

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