Pada suhu 27°C glukosa (mr=180) sebanyak 18 gram dilarutkan dalam air sampe volumenya 400 ml,tentukan tekanan osmotik larutan

T = 27°C = 300K
Mr = 180
wt = 18 gram
V = 400 mL
π = ................................?

π = M x R x T
π = wt/Mr x 1000/V x 300
= 18/180 x 1000/400 x 300
= 0,1 x 750
= 75 atm