Question

18 g of glucose is added to 178.2 g of water vapour presuure is?

Answer 1

molecular mass of water is 18g    

for 178.2  g   then 178.2/18 =9.9 n(a)

molecular mass of glucose are 180 g then n(b) = 0.1

where xb=0.1/0.1+9.9=0.01

x(a) =0.01

according to the roults law for non voletile solute.......

p=poAXoA =PoA(1-XB)

p=760(1-0.01)

760-7.6=752.4