Question
Asked Jan 31, 2020
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A certain liquid X has a normal boiling point of 90.80°C and a boiling point elevation constant =Kb 2.32·°C·kg/mol−1. Calculate the boiling point of a solution made of 7.14g of ammonium sulfate (NH42SO4) dissolved in 150.g of X.

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Expert Answer

Step 1

Boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the atmospheric pressure. As the vapor pressure of a solution containing non-volatile solute is lower than that of the pure solvent, its boiling point will be higher than that of the pure solvent. The increase in boiling point is called as elevation in the boiling point. This elevation is represented as ∆Tb. If Tb is the boiling point of the solution, T0b is the boiling point of pure solvent, Kb is the molal elevation constant or ebullioscopic constant and m is the molality of the solution then elevation in boiling point can be written as-

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Step 2

Ammonium sulphate is the solute dissolved in the liquid X. The molar mass of ammonium sulphate MA= 132.14 g. The given mass of ammonium sulphate WA is 7.14 g. let the number of moles of ammonium sulphate be nA. The value of number of moles of ammonium sulphate can be calculated as follows-

Chemistry homework question answer, step 2, image 1
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Step 3

The weight of the solvent WB= 150g=0.15kg. The molality of the solution can be calculated by dividing t...

Chemistry homework question answer, step 3, image 1
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