What is the vapor pressure at 23 oc of a solution of 1.20 g of naphthalene, c10h8, in 25.6 g of benzene, c6h6? . the vapor pressure of pure benzene at 23 oc is 86.0 mm hg; the vapor pressure of naphthalene can be neglected. calculate the vapor-pressure lowering of the solution.

The vapor-pressure lowering of the solution is 2,39 mmHg

Explanation:

The Raoult's law says that the vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent. The vapor pressure lowering is directly proportional to the mole fraction of the solute. In mathematical terms that means:

ΔP = (1)

Where:

ΔP is the vapor-pressure lowering of the solution

is molar fraction of solute

And is the capor pressure of the pure solvent (86,0 mmHg)

The moles of naphthalene are:

9,36x10⁻³ moles solute

And the moles of benzene are:

0,328 moles solvent

Molar fraction of solute is:

0,0278

Replacing in (1)

ΔP = 2,39 mm Hg

I hope it helps!

the answer would be b i believe

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.