Question

Sebanyak 171 gram Sukrosa dilarutkan dalam X gram air bila diketahui harga tekanan uap air murni 22,5 mmHg dan tekanan uap larutan sukrosa 20,5 mmHg maka tentukan Harga X?

Answer 1

Penjelasan:

P larutan = X pelarut.P air murni

20,5 = X pelarut.22,5

X pelarut = 0,91

Mr sucrose = 342 g/mol

mol sucrose = 171÷342 = 0,5 mol

mol total = mol air + mol sucrose

X pelarut = mol air÷mol total

X pelarut = mol air÷(0,5+ mol air)

0,91 = mol air÷(0,5+mol air)

0,455+0,91 mol air = mol air

mol air = 5,06

Mr air = 18 g/mol

massa air = mol air×Mr air = 5,06×18 = 91,08 gram

Answer 2

Jawaban:

93,258 gram

Penjelasan:

diketahui :

massa sukrosa (C12H22O11) = 171 gram

Ppel = 22,5 mmHg

Plar = 20,5 mmHg

ditanya : massa H2O?

jawab :

* Mr sukrosa = 12(12) + 1(22) + 16(11) = 342

  Mr H2O = 18

* Plar = Ppel - delta p

  20,5 mmHg = 22,5 mmHg - delta p

   2 = delta p

* delta p = Ppel . Xter

  2 mmHg = 22,5 mmHg . Xter

  0.088 = Xter

* Xter = mol ter/mol ter + mol pel

  0.088 = 0,5/n+0,5

   0.088(n+0,5) = 0,5

    n =5,181 mol

* mol = gram/Mr

 5,181 mol = gram/18

  93,258 gram = massa H2O