The contents of the beaker shown below were the result of thoroughly mixing two

ID: 1043044 • Letter: T

Question

The contents of the beaker shown below were the result of thoroughly mixing two liquids and then letting the beaker stand undisturbed.

The two liquids were methanol (CH3OH, d = 0.787 g/mL) and toluene (C6H5CH3 d = 0.866 g/mL).

Question 10 options:

These two liquids are immiscible with methanol forming the bottom layer.

These two liquids are miscible with methanol forming the top layer.

These two liquids are immiscible with toluene forming the bottom layer.

These two liquids are miscible with toluene forming the top layer.

These two liquids are immiscible with methanol forming the bottom layer.

These two liquids are miscible with methanol forming the top layer.

These two liquids are immiscible with toluene forming the bottom layer.

These two liquids are miscible with toluene forming the top layer.

here in CH3OH is polar solevent because the hydroxyl group (O-H) is polar and whole molecule is non -symmetrical.

toulene is nonpolar solvent because it is hydrocarbon with congugation( double bonds)

when we mixed these two solvents these are immscible becasue polar and nonpolar solvents doesnot miscible with each other at normal conditions.

the solvent with high density will remains at bootom layer and lesserw ill float.

hence the correct option is : C. two liquids are immiscible with toulene as bottom layer.

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.