Chemistry, asked by pabhishek8187, 1 year ago

The mixture of immiscible liquid and water boils at 98°C at 755 mm Hg. The vapour pressure of water at this temperature is 712 mm Hg. Find the weight composition of the distillate. [Given : Mol. wt. of immiscible liquid = 204]

Answers

Answered by Anonymous
1

Oil and water are examples of immiscible liquids - one floats on top … ... The vapor pressure of mixtures of immiscible liquids ... would boil at a temperature just a shade less than 98°C - in other ...

Answered by anvitanvar032
0

Answer:

The correct answer of this question is 44.3%

Explanation:

Given - The mixture of immiscible liquid and water boils at 98°C at 755 mm Hg. The vapor pressure of water at this temperature is 712 mm Hg.

To Find - Find the weight composition of the distillate.

The optimal distillate-rate strategy, D(t), is defined as the policy that delivers the largest amount of distillate product of a certain purity, y*, in a given operating time, T.

P_{T} =P^{0} _{A} + P^{0} _{B}

Two immiscible liquid = 755 - 712 = 43

\frac{w_{A} }{W_{B} } = \frac{P^{0}_{A} m_{A}   }{P^{0} _{B} m_{B}  }

= 60 × 93 / 700 × 18 = 44.3%

#SPJ2

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