Question

Tekanan uap jenuh air pada 100°c adalah 760 mmhg berapakah tekanan uap jenuh larutan glukosa 18% pada 100°C (mr =glukosa 180)

Answer 1

Po= 760
m glukosa = 18/100 × 100= 18 gr
m air = 72/100 ×100 = 72 gr
P= Xb × Po
Xb = nb/ nb + na
nb = mb/mr = 18/180 = 0,1
na = ma/mr = 72/18 = 4
xb= 4/4+0,1 = 4/4,01 = 0,998
P = xb ×Po
P = 0,998×760
P= 758,48