A solution contains 57.0 g octane, C_8 H_18 (M = 114 g/mol) dissolved in 148 g o

Question

A solution contains 57.0 g octane, C_8 H_18 (M = 114 g/mol) dissolved in 148 g of benzene, C_4 H_6 (M = octane is a nonelectrolyte, the freezing point of pure C_6 H_6 is 5.50 degree; K_f for octane is 3.47 degree C/molal and the K_f for C_6 H_6 is 5.12 degree C/molal. The density of the solution is 0.83 g/mL. The molality of C_6 H_16 in the solution is m. a) 1.30 b) 0.500 c) 0.296 d) 3.38 e) 0.00338 The molarity of the solution is M. a) 2.02 b) 0.494 c) 0.942 d) 2.80 e) indicate some other answer The freezing point of the solution is degree C. a) -1.18 b) -6.22 c) 2.94 d) -17.3 e) indicate some other answer degree C

Explanation / Answer

1. molality of mixer(m) = n/W in kg

   n = no of mole of solute(octane) = w/Mwt = 57/114 = 0.5 mole

W = mass of benzene(solvent) in kg = 0.148 kg

m = 0.5/0.148 = 3.38 molal

answer: d

2. molarity of solution = n/V

n = no of mole of solute(octane) = w/Mwt = 57/114 = 0.5 mole

v = volume of solution in L

mass of solution = 57+148 = 205 g

volume of solution = mass/density = 205/0.83 = 247 ml

molarity(M) = 0.5/0.247 = 2.02 M

answer: 2.02 M

3. DTf = i*kf*m

(T0 - Ts) = i*kf*m

i = vanthoff factor = 1

kf of benzene = 5.12 c/molal

m = molality of mixer = n/W in kg = 3.38 m

T0 = freezing point of solvent(C6H6) = 5.5 c

Ts = freezing point of solution = ?

(5.5-x) = 1*5.12*3.38

x = -11.8 C

freezing point of solution = -11.8 C

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