Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

At ${{25}^{\circ }}C$ benzene and toluene have densities 0.879 and 0.867 g/mL respectively. Assuming that benzene, toluene solutions are ideal, then the equation for the density (e) of solution as: $e=\dfrac{1}{100}\left[ 0.879V + 0.867(100 - V) \right]$, where V is the volume of benzene.

seo-qna
Last updated date: 27th Jul 2024
Total views: 405.3k
Views today: 7.05k
Answer
VerifiedVerified
405.3k+ views
Hint: For solving this question:
-Find the mass of benzene and toluene as density is given using the formula: mass= density $\times $ volume,
- then find the density of the solution.

Complete step by step solution:
We have been provided that the densities of benzene and toluene are 0.879 and 0.867 g/mol,
As benzene and toluene form an ideal solution,
An ideal solution is a mixture in which the molecules of different species are distinguishable, however, unlike the ideal gas, the molecules in an ideal solution exert forces on one another. When those forces are the same for all molecules independent of species then a solution is said to be ideal.
So, the volume of the solution would be:
\[{{V}_{sol}} = {{V}_{ben}} + {{V}_{tol}}\]
Now, we will be calculating the mass as we have been provided with density of benzene and toluene,
For that let the volume of toluene and benzene as ${{V}_{tol}}$ and ${{V}_{ben}}$,
Mass of toluene: $0.867\times {{V}_{tol}}$,
Mass of benzene: $0.879\times {{V}_{ben}}$,
So, the total mass of solution would be: $0.867\times {{V}_{tol}}$ + $0.879\times {{V}_{ben}}$,
Hence, the density of solution would be: total mass/ total volume,
Keeping the values in the above formula: $e = \dfrac{0.867\times {{V}_{tol}} + 0.879\times {{V}_{ben}}}{{{V}_{ben}}+{{V}_{tol}}}$:
Hence, we can conclude that the statement is incorrect.

Note: Note that, there is an another method to determine the density of binary mixture .the method involves the use of mole fraction as:
$\text{ }{{\text{D}}_{\text{mix}}}\text{ = }\dfrac{100}{\left( {{\text{X}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{\text{D}}_{\text{1}}} \right)\text{ + }\left( {{\text{X}}_{\text{2}}}\text{ }\!\!\times\!\!\text{ }{{\text{D}}_{\text{2}}} \right)\text{ }}\text{ }$
Where ,${{\text{X}}_{\text{1}}}$ is the mole fraction of component 1 and ${{\text{X}}_{2}}$ is the mole fraction of second component.