Question

Asked Sep 4, 2019

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A solution of 0.750 grams of a non electrolyte in 40.0 grams of CCl_{4 }boils at 78.6°C. Calculate the molecular weight of the solute if the boiling point constant for CCl_{4} is 5.02°C and the boiling point of the pure solvent is 76.8°C.

Step 1

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

∆T_{b }= T_{b}-T_{b}^{○} …… (1)

Here, ∆T_{b } is the elevation in boiling point, T_{b }is the boiling point of the solution and T_{b}^{○ } is the boiling point of pure solvent.

Elevation in boiling point is also represented as,

∆T_{b }= K_{b}×m …… (2)

Here, ∆T_{b } is the elevation in boiling point, K_{b }is the boiling point constant and m is the molality.

Molality is expressed with the help of equation (3) in which W_{2} is the weight of the solute, is the molecular weight of solute and W_{1} is the weight of solvent.

Step 2

The elevation in boiling point is calculated by using equation (1) and substituting 78.6^{○}C for T_{b} and 76.8^{○}C for T_{b}^{○}.

∆T_{b }= 78.6^{○}C -76.8^{○}C

=1.8^{○}C

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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