Question

115 grams of KCl is dissolved in 750 ml of water (assume density = 1.005 g/ml). What are the molality, molarity, mole fraction, mole percent, % mass, ppm by mass? What would be the freezing point and boiling point of that solution assuming the Kf of water is 1.86 oC/m and Kb is 0.512 oC/m (assume that KCl fully dissociates with no pairing of ions)?

Answer 1

Question #80828, Chemistry / General Chemistry | for completion

115 grams of KCl is dissolved in 750 ml of water (assume density = 1.005 g/ml). What are the molality, molarity, mole fraction, mole percent, % mass, ppm by mass? What would be the freezing point and boiling point of that solution assuming the Kf of water is 1.86 oC/m and Kb is 0.512 oC/m (assume that KCl fully dissociates with no pairing of ions)?

m(KCl) = 115g

V(H2O) = 750 ml

p(solution) = 1,005g/ml

Cm-?

Mu -?

X-?

W-?

Kf = 1.86 oC/m

Kb = 0.512 oC/m

Solution:

m(H2O) = 750 g

n(KCl) = m(KCl)/M(KCl) = 115 / 74 = 1.55 mol

n(H2O) = m(H2O)/M(H2O) = 750 / 18 = 41.6 mol

X(KCl) = n(KCl) / (n(KCl) + n(H2O)) = 1.55 / (1.55 + 41.6) = 1.55 / 43.15 = 0.036 = 3.6%

Mu = n/m(solvent) = 1.55 / 750 = 0.002 mol/g = 2 mol/kg

Cm = n/V(solution)

V = m/p = (750 + 115) / 1.005 = 860 ml = 0.86 l

Cm = 1.55 / 0.86 = 1.8 mol/l

W = m(soluted) / m(solution) = 115 / (750 + 115) = 0.132 = 13.2%

ΔT = k*Mu

ΔTf = Kf*Mu = 1.86*2 = 3.75

ΔTb = Kb*Mu = 0.512*2 = 1.024

Answer: Cm = 1.8 mol/l, Mu = 2 mol/kg, x = 0.036 = 3.6%, w = 13.3%, ΔTf = 3.75, ΔTb = 1,024

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