tentukanlah tekanan osmotik larutan yang tiap liternya mengandung 0,6 gr urea pada suhu 25° C ( mr urea = 60 )

diket :
massa urea = 0.6 gram
mr urwa = 60
vol = 1 L
T = 25°C = 25+273=298 K
R = 0.082

dit:π???
jawab..
π = MxRxT

M = n/v

n = massa/mr = 0.6/60= 0.01 mol

M = 0.01 mol/1L = 0.01 M

π=MxRxT
π=0.01x0.082x298
π=0.24436 atm