Question

1. Tekanan uap air pd suhu 25 drjt C adl 40 mmHg . Hitung tekanan larutan glukosa 10%
2. 10gr zat Y dilarutkan dlm 180gr air pd suhu 25 drjtC. Ternyata tkanan uap larutan turun 0,2 mmHg. Tekanan uap murni 25,20 mmHg. Tent massa molekul relatif zat Y

Answer 1

1. diketahui: P^o= 40mmHg
                   glukosa 10 %
  dit: P?
  massa pelarut= 100%-10%= 90%
mol air(pelarut) = gr/Mr
                       = 90/18 = 5 mol
mol glukosa (terlarut) = gr/Mr
                                = 10/170
                                = 0.058 mol
Xp= mol pelarut/mol pelarut+mol terlarut
    = 5/5+0.058
    = 5/5,058
    = 0,98
P=Po . Xp
  = 40 . 0,98
  = 39,2 mmHg  

2. 10 gr zat Y
    180 gr air
     P turun 0,2 mmHg
    Po= 25,20 mmHg
    Mr?
    P= 25,20 - 0,2 = 25 mmHg
     mol air = gr/Mr = 180/18 = 10 mol
   P = Po . Xp
   P= Po (mol p / mol p+mol t)
  25= 25,20 (10 / 10 + 10/Mr)
  25 = 25,20 (10 / 10Mr/Mr+10/Mr)
  25 = 25,20 (10Mr / 10Mr+10)
  25 = 252Mr / 10Mr+10
  250Mr+250 = 252Mr
  250 = 252Mr - 250Mr
  250 = 2Mr
  Mr = 125