Asked Sep 4, 2019

A solution of 0.750 grams  of a non electrolyte in 40.0 grams of CClboils at 78.6°C. Calculate the molecular weight of the solute if the boiling point constant for CCl4 is 5.02°C and the boiling point of the pure solvent is 76.8°C.


Expert Answer

Step 1

Elevation in boiling point is calculated with the equation (1) given below.

∆Tb = Tb-Tb                                                                                                                  …… (1)

Here, ∆Tb  is the elevation in boiling point, Tb is the boiling point of the solution and Tb is the boiling point of pure solvent.

Elevation in boiling point is also represented as,

∆Tb = Kb×m                                                                                                                  …… (2)

Here, ∆Tb  is the elevation in boiling point, Kb is the boiling point constant and m is the molality.

Molality is expressed with the help of equation (3) in which W2 is the weight of the solute, is the molecular weight of solute and W1 is the weight of solvent.

Wx 1000

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Wx 1000 m M,xW (3)

Step 2

The elevation in boiling point is calculated by using equation (1) and substituting 78.6C for Tb and 76.8C for Tb.

∆Tb = 78.6C -76.8C


Step 3

Furthermore, the molality is calculated by rearranging the equation (2) in terms of molality as,

m= ∆Tb /Kb

Substitute 1.8○C for ∆Tb  and 5.02○Cm-1 for Kb

m= 1.8○C/5.02○Cm-1


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