Sebanyak 18 gram glukosa (M=180) dilarutkan dalam 90 gram (M=18).jika diketahui tekanan uap jenuh air pada suhu 100°C adalah 760 mmHg,maka besarnya penurunan te

dik: masaa glukosa :18g
massa H2O. :90g
mr glukosa :180
mr air. :18
P°. :760mmhg
dit: P?
jawab :
xpelarut = mol H2O/molH2O+mol gluko
= 90/18 ÷ 90/18+18/180
= 5 ÷ 5+0,1
= 5 ÷ 5,1
= 0,98
P =xH2O × P°
=0,98×760
=744,8 mmHg

wt = 18 gram (Mr = 180)
wp = 90 gram (Mr = 18)
P = 760 mmhg
T = 100°C
ΔP = ................... ?

nt = wt/Mr = 18/180 = 0,1 mol
np = wp/Mr = 90/18 = 5 mol

Xp = np/(np+nt) = 5/(5+0,1) = 5/5,1 = 0,98
Xt = 1 - 0,98 = 0,02
 
ΔP =Xt x P°
= 0,02 x 760
= 15,2 mmHg