   Chapter 11, Problem 131IP

Chapter
Section
Textbook Problem

# An aqueous solution containing 0.250 mole of Q, a strong electrolyte, in 5.00 × 102 g water freezes at −2.79°C. What is the van’t Hoff factor for Q? The molal freezing-point depression constant for water is 1.86°C · kg/mol. What is the formula of Q if it is 38.68% chlorine by mass and there are twice as many anions as cations in one formula unit of Q?

Interpretation Introduction

Interpretation: The formula for Q and the Van’t Hoff factor for Q has to be determined.

Concept Introduction:

The changes in boiling and freezing point are by given by the modified equation that is used to determine the colligative properties of strong electrolytes. The van’t Hoff factor in the equation is given by,

ΔT=imK

Where, K= change in boiling/freezing point

i=Van’t Hoff factor that takes the equation,

i=molesofparticlesinsolutionmolesofsolutedissolved

Molar mass of a compound can be given by atomic mass of a element to the total mass of the compound. It can be given by expression

Molarmass=MassofaelementTotalmass

Explanation

Moles of Q                                           = 0.250mole

Mass of Q                                             = 5.00×102g

Freezing point of Water     = -2.79°C

Molal freezing point depression constant = 1.86°C.kgmol-1

To calculate the Van’t Hoff factor for Q

ΔT=imKf

Van’t Hoff factor for Q , i= ΔTmKf

= 2.79°C0.250mol0.500kg×1.86°Ckgmol

= 3.00

Van’t Hoff factor for Q = 3.00

Record the given info

Mass percent of chlorine = 38.68%

To find the formula unit for Q

van’t Hoff factor for Q = 3.00

The formula unit for Q is MCl2

To calculate the moles of ClandM

Atomic mass of chlorine = 35

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