Information that may be useful: N_A = 6.022 times 10^23 delta G degree = delta H

Question

Information that may be useful: N_A = 6.022 times 10^23 delta G degree = delta H degree - T delta S degree T(K) = T(degree C) + 273 delta T_b = K_b m I delta T_f = K_f m i Replacement question #5: An unknown alcohol (that is non-volatile and a non-electrolyte) obtained from a tropical plant has been found useful in the household cleaning industry. A solution of 15.0 g of the alcohol in 0.150 kg of benzene freezes at 2.76 degree C. What is the approximate molar mass (g/mol) of this alcohol? Benzene: normal freezing point is 5.50 degree C and K_f = 5.12 C/m

Explanation / Answer

information given

Solute present in the solution = 15.09 g = 0.015 Kg

Amount of solution = 0.150 Kg

Freezing point of solvent = 5.50 C

Freezing point of solution after adding solute = 2.76 Degree celsius

Solution :

Given problem is related to colligative properties ( Depression in freezing point)

Tf = Kf × b × i

Tf = 5.50 C - 2.76 C (difference in temperature)

Kf = 5.12 C Kg/mol = cryoscopic constant

b = molality = (mass of solute/molecular mass of solute)/(mass of solvent)

= (0.015 Kg / Msolute)/(0.150 Kg)

i = Van't hoff factor = 1 ( for alcohols, because they dont split in benzene)

Hence

2.74 = 5.12 × (0.015/Msolute)/(0.150)

Msolute = 0.1868 Kg mol-1

= 186.8 g mol-1

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