2,4 gram urea yang dilarutkan dalam air sampai volume 250 ml pada suhu 27 ⁰C .tentukan tekanan osmotiknya
tolong ya pr buat besok nih
AndreasWatang
Mr Urea = 60 n = g / Mr = 2.4 / 60 = 0.04 mol M = n/V = 0.04 / 0.25 = 0.16 M T = (27+273)K = 300K Phi = M R T = 0.16 M × 0.082 atm L / mol K × 300 K = 3.936 atm
n = g / Mr
= 2.4 / 60
= 0.04 mol
M = n/V
= 0.04 / 0.25
= 0.16 M
T = (27+273)K
= 300K
Phi = M R T
= 0.16 M × 0.082 atm L / mol K × 300 K
= 3.936 atm