Question

What mass of glucose (in grams) is added to 72 g of water which lower vapour pressure by 2%? ​

Answer 1

Answer:

H2NCONH2)→M=60g/mol

m= molar mass

MC6H12O6=12×6+12×1+6×6=72+12+96+180g/mol

Relative lowering in Vapour pressure=no.ofmoles×massofwatermolarmassofsolvent=0.33×5018

180W×10018=601×5018W=6g

Answer 2

Given:

Mass of water = 72 gm

Lowering of vapour pressure = 2%

To Find:

The mass of glucose required.

Calculation:

- The moles of water present in 72 gm= 72 / 18 = 4

- Let no of moles of glucose be n.

- We know that relative lowering of vapour pressure is equal to the mole fraction of the solute dissolved.

⇒ Relative Lowering of vapour pressure of the given solution = Mole fraction of glucose

⇒ 2/100 = n/(n+4)

⇒ n = 4/49

⇒ n = 0.0816

- Mass of glucose = 0.0816 × 180

⇒ M = 15.5 gm

- So the required mass of glucose is 15.5 gm.