Question

In a liquid mixture of chloroform (C) and acetone (A) with chloroform mole fraction xC = 0.4693, the partial molar volumes are 80.24 mL for chloroform and 74.17 mL for acetone. Find the volume occupied by 1000 g of this mixture.

Answer 1

Answer:

The volume occupied by the 1000 gm of this mixture is V = 886.82 $cm^{3}$

Explanation:

Given data

Denote

Chloroform = C and Acetone = A

No. of moles of Acetone = $n_{A}$

No. of moles of Chloroform = $n_{c}$

MW of acetone = 58.08 gm $mol^{-1}$

MW of chloroform = 119.37 gm $mol^{-1}$

Volume of chloroform = 80.24 ml

Volume of acetone = 74.17 ml

The total number of moles present can be calculated as

1000 g = $n_{A}$ × 58.08 + $n_{c}$ × 119.37

1000 g = n (1- $x_{c}$) × 58.08 + n $x_{c}$ × 119.37

Since $x_{c} = 0.4693$

1000 = n (1-0.4693) × 58.08 + n × 0.4693 × 119.37

1000 = n × 0.5307 × 58.08 +  n × 56.02

n = 11.515 moles

Therefore $n_{A}$ = (1-0.4693) 11.515 = 6.111 moles

$n_{c}$ = 0.4693 × 11.515 = 5.404 moles

Thus the total volume of the solution is given by

V = $n_{A}$ $V_{A}$ + $n_{c}$ $V_{C}$

Put all the values in above formula we get

V = 6.111 × 74.166 + 5.404 × 80.235

V = 886.82 $cm^{3}$

Therefore the volume occupied by the 1000 gm of this mixture is V = 886.82 $cm^{3}$