Question

Jika diketahui kb air=0,52, Mr Na2SO4=142 dan massa jenis larutan Na2SO4 0,04M = 1,1
Tentukan titik didih 3L larutan Na2SO4

Answer 1

Massa jenisnya mungkin gram/mL ya ._.

volume = 3 liter = 3000 mL
massa jenis = 1,1 gram/mL
Kb = 0,52
M = 0,04 M
Mr Na2SO4 = 142
Tb° = 100°C
i = 3 (asumsi Na2SO4 mengion sempurna)
Tb = ?

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mol Na2SO4 = M × v = 0,04 × 3 = 0,12 mol
massa Na2SO4 = mol × Mr = 0,12 × 142 = 17,04 gram

massa larutan = massa jenis × volume = 1,1 × 3000 = 3300 gram
massa air = massa larutan - massa Na2SO4 = 3300 - 17,04 = 3282,96 gram

molalitas = mol Na2SO4 / massa air (kg)
molalitas = 0,12 / 3,28296
molalitas = 0,0365

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∆Tb = m × Kb × i
∆Tb = 0,0365 × 0,52 × 3
∆Tb = 0.05694

Tb = 100 - 0.05694 = 99.94306°C