The vapour pressure of water at 293 K is 2.32 kPa. When 16.8 g of glucose (molar mass 180.15 g/mol) is dissolv

Question

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SORRY!! Question is supposed to read:
The vapour pressure of water at 293 K is 2.32 kPa. When 16.8 g of glucose (molar mass 180.15 g/mol) is dissolved in 100 mL of water, the vapour pressure is...

Answer 1

p(a) = p° (a) x a

16.8 /180.15 = 0.0932 moles glucose

for the water 100 mL = 100 g

100 g / 18 g/mol = 5.55 moles H2O

xa = 5.55 / 5.55 + 0.0932 = 0.982

p(a) = 2.32 x 0.982 = 2.28 kPa

Answer 2

2.275 kPa.