Latihan dan Jawaban Soal Sifat Koligatif Larutan dan Satuan Konsentrasi Larutan

Latihan Soal Sifat Koligatif Larutan dan Satuan Konsentrasi Larutan

1.           Berapa gram kristal H₂C₂O₄ ∙ 2H₂O yang harus dilarutkan ke dalam air agar didapatkan 600 mL larutan H₂C₂O₄ 0,5 M? (Ar C = 12, O = 16, H = 1)

2.           Di dalam 400 mL larutan amonia, terlarut 3,4 gram NH₃ ∙ (Ar N = 14, H = 1). Hitunglah :

              a.          molaritas larutan

              b.         fraksi mol larutan

3.           Hitunglah molalitas dan fraksi mol larutan NaOH dalam air yang kadarnya 40% (M_r NaOH = 40)

4.           Jika perubahan volume dalam pelarutan diabaikan, tentukan molaritas dan molalitas larutan yang terjadi jika 100 mL alkohol dicampur dengan 200 mL air (massa jenis alkohol dan air dianggap sama dengan 1 gram/mL).

5.           Berapa gram KOH yang harus dilarutkan ke dalam 400 mL air agar didapatkan larutan KOH 0,2 M? (Ar K = 39, O = 16, H = 1)

Jawaban Soal Sifat Koligatif Larutan dan Satuan Konsentrasi Larutan

1.           Mr H₂C₂O₄                        = 90

              V                                 = 0,6 L

              M                                 = 0,5 M

              Mr H₂C₂O₄ ∙ 2H₂O      = 90+36=126 gram/mol

              n                                  = M x V

                                                   = 0,5 x 0,6

                                                   = 0,3 mol

              masa Kristal H₂C₂O₄ ∙ 2 H₂O :

              gr                                 = n x Mr

                                                   = 0,3 x 126

                                                   = 37,8 gram

·         Zat terlarut             à        Volume sedikit

·         Pelarut                   à        Volume banyak

·         Pelarut biasanya air

2.         Mr                              = 17

            n                                  = 3,4/17

                                                = 0,2

            a.         M                                 = 0,2/0,4

                                                            =0,5

                        Masa Pelarut                = 400 gram – 3,4 gram

                                                            =396,6 gram

            b.         nH₂O                           = 396,6/18

                                                            = 22,03

                        Fraksi mol NH₃           = 0,2/22,23

                                                            = 0,009

3.         masa NaOH                = 40% x 100

                                                = 40 gram

            masa air                       = 100 – 40

                                                = 60 gram

            nNaOH                       = 40/40

                                                = 1

            m                                 = n x 1000/p

                                                = 1 x 1000/60

                                                = 16,67 molal

            nH₂O                           = 60/18

                                                =3,33

            X                                 = 1 / 4,33

                                                = 0,23

4.         V                                 = 100+200

                                                = 300 mL à 0,3 L

            n                                  = 100 /46

                                                = 2,17

            M                                 = 2,17/0,3

                                                =7,23 mol/L

            m                                 = 2,17 x 1000/200

                                                = 10,85

5.         n                                  = V x M

                                                = 0,4 x 0,2

                                                = 0,08

            0,08                             = gr/56

            gr                                 = 4,48 gram