A). 1.43 grams of an organic solute dissolved in 50.0 grams of benzene has a fre

Question

A). 1.43 grams of an organic solute dissolved in 50.0 grams of benzene has a freezing point of 4.2 C. What is the molecular weight of the solute?

B. An organic compound whose molecular weight is 140.0 grams/mol is dissolved 1000.0 grams of cyclohexane. The freezing point of the solution is 4.75 C. How much of the compound was dissolved in the cyclohexane?

C. What would be the mass needed to make 350.0g of a water and calcium chloride that would lower the freezing point of water to -5.85C?

D. What would be the mass needed to make the above solution boil at 110.6C?

E. What quantity in grams of methanol, CH3OH is required to prepare a 0.244 m solution in 400.0 grams of water.

Explanation / Answer

a) freezing point of pure benzene = 5.5 0C

therefore depression in freezing point = 5.5-4.2 = 1.3 0C

now, depression in freezing point = molality*Kf

now, molal depression constant for benzene, Kf = 5.12 0C/m

thus, 1.3 = molality*5.12

or, molality = 0.254 m

now, molality,m = moles of solute/mass of solvent in kg

thus, moles of solute = 0.254*0.05 =0.0127

now, moles of solute = mass of solute/molecular mass of solute

or, molecular mass of solute = 112.64 g

B) freezing point of cyclohexane = 6.5 0C

and molal depression constant for cyclohexane = 20.2 0C/m

thus, depression in freezing point = 6.5-4.75 = 1.75

or, 1.75 = 20.2*m

or, molality,m = 0.0866

thus, moles of solute = 0.0866*1 = 0.0866

thus, mass of compound dissolved = 0.0866*molecular mass = 0.0866*140 = 12.13 g

C)  freezing point of water = 0 C

depression in freezing point = 5.85

molal depression constant for water = 1.52

thus, molality = 5.85/1.86 = 3.15 m

moles of CaCl2 reqquired = molality*mass of water in kg = 1.101

mass of CaCl2 required = moles of CaCl2 * molecular mass of CaCl2 = 1.01*111 = 122.2 g

D) boiling point of pure water = 100 C

elevation constant for water = 0.52

elevation in boiling point = 10.6 C

now, elevation in boiling point = Kb*molality = 0.52*m

or, 10.6 = 0.52*m

or, m = 20.385

moles of CaCl2 required = 20.385*0.35 = 7.135

mass of CaCl2 required = 7.135*111 = 791.94 g

E) molecular mass of CH3OH = 32 g

molality = moles of CH3OH/mass of water in kg

thus, moles of CH3OH = 0.244*0.4 = 0.0976

mass of CH3OH required = moles*molecular mass = 3.1232 g

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