Some nonelectrolyte solute (molar mass = 142 g/mol) was dissolved in 150. mL of a solvent (density = 0.879 g/cm 3 ). The elevated boiling point of the solution was 355.4 K. What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 kJ/mol, the entropy of vaporization is 95.95 J/K mol, and the boiling-point elevation constant is 2.5 K kg/mol.

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Interpretation Introduction Interpretation: The molar mass of solute, volume, density, enthalpy of vaporization, entropy of vaporization and boiling point elevation constant of solvent is given. The mass of solute that was dissolved in the solvent is to be calculated. Concept introduction: The expression of elevation in boiling point is, Δ T = 1000 × k b × ω m × W Explanation Given Enthalpy of vaporization of solvent is 33.90 kJ/mol . Entropy of vaporization of solvent is 95.95 J/K ⋅ mol The conversion of kilo-joule ( kJ ) into joule ( J ) is done as, 1 kJ = 10 3 J Hence, The conversion of 33.90 kJ into joule is, 33 .90 kJ = ( 33 .90 × 10 3 ) J = 33 .90 × 10 3 J Formula The entropy of vaporization is calculated as, Δ S = Δ H vap T Where, Δ H vap is the enthalpy of vaporization. Δ S is the entropy of vaporization. T is the elevation in boiling point of solvent. Substitute the values of Δ H vap and Δ S in the above equation. Δ S = Δ H vap T T = ( 33 .90 × 10 3 95.95 ) K = 3 5 3 . 3 0 K _ The elevation in boiling point of solvent is 353.3 K . The elevated boiling point of solution is 355.4 K . The expression of boiling point elevation is calculated as, Δ T = T 1 − T Where, Δ T is the boiling point elevation. T 1 is the elevated boiling point of solution. Substitute the values of T 1 and T in the above equation. Δ T = T 1 − T = 355.4 K − 353

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Interpretation Introduction

Interpretation: The molar mass of solute, volume, density, enthalpy of vaporization, entropy of vaporization and boiling point elevation constant of solvent is given. The mass of solute that was dissolved in the solvent is to be calculated.

Concept introduction: The expression of elevation in boiling point is,

ΔT=1000×kb×ωm×W

Explanation of Solution

Given

Enthalpy of vaporization of solvent is 33.90 kJ/mol .

Entropy of vaporization of solvent is 95.95 J/K⋅mol

The conversion of kilo-joule (kJ) into joule (J) is done as,

1 kJ=103 J

Hence,

The conversion of 33.90 kJ into joule is,

33.90 kJ=(33.90×103) J=33.90 ×103 J

Formula

The entropy of vaporization is calculated as,

ΔS=ΔHvapT

Where,

ΔHvap is the enthalpy of vaporization.
ΔS is the entropy of vaporization.
T is the elevation in boiling point of solvent.

Substitute the values of ΔHvap and ΔS in the above equation.

ΔS=ΔHvapTT=(33.90 ×103 95.95) K=353.30 K_

The elevation in boiling point of solvent is 353.3 K .

The elevated boiling point of solution is 355.4 K .

The expression of boiling point elevation is calculated as,

ΔT=T1−T

Where,

ΔT is the boiling point elevation.
T1 is the elevated boiling point of solution.

Substitute the values of T1 and T in the above equation.

ΔT=T1−T=355.4 K−353

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