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

The volume occupied by the 1000 gm of this mixture is V = 886.82
`cm^(3)`

Step-by-step 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)`