Question

What is the molality of a solution prepared by dissolving 0.114 mol of chloroform, CHCl₃, in 476 g of toluene, C₆H₅CH₃?

a,0.543 m
b.0.0216 m
c.0.0220 m
d,0.0287 m
e,0.239 m

Answer 1

The molality of a solution prepared by dissolving 0.114 mol of chloroform in 476 g of toluene is calculated by dividing the moles of solute by kilograms of solvent, yielding a molality of 0.239 m.

Step-by-step explanation:

To calculate the molality of a solution prepared by dissolving 0.114 mol of chloroform (CHCl₃) in 476 g of toluene (C₆H₅CH₃), we use the formula:

Molality (m) = moles of solute / kilograms of solvent

First, convert the mass of toluene from grams to kilograms:

476 g = 0.476 kg

Next, plug the values into the formula:

Molality (m) = 0.114 mol / 0.476 kg = 0.239 m

This indicates that the correct answer is (e) 0.239 m, which represents the molality of the chloroform solution in toluene.