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