Question

Question Part A A solution is prepared by dissolving 23.7 g of (111.0 g/mol) in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of in this solution is ________ molarity

Part B The freezing point of benzene C6H6 is 5.5000C at 1 atmosphere. In a laboratory experiment, students synthesized a new compound and found that when 12.70 grams of the compound were dissolved in 274.0 grams of benzene, the solution began to freeze at 4.4550C. The compound was also found to be nonvolatile and a non-electrolyte. Kb (0C / m) = -5.12 What is the molecular weight they determined for this compound ?________________g/mol

Answer 1

Part A : Molarity = moles of solute/ liter of solution

mass of solution= mass of water+ mass of substance= 375+23.7=398.7 gm

Density of the solution = 1.05 g/mol, so volume of the solution= 398.7/1.05 g/ml= 379.7 ml =0.3797L

moles of the substance= mass/molecular weight= 23.7/111 =0.2135 moles

Molarity= moles of solute/L of the solution = 0.2135/0.3797= 0.5623M

Part B:

depresion in freezing point = kf*m, m= molality= moles of solute/ kg of solvent

Depressin in freezing point = 4.455-5.5=-1.045 deg.c

-1.045= -5.12*m

m= 1.045/5.12=0.2041

Moles of solute= mass/Molecular weight= 12.7/M, M= Molecular weight

Mass of solvent in Kg= 274 gm =0.274Kg

1 kg contains 0.2041 moles of solute

0.274 kg contains 0.274*0.2041 moles of solute =0.055923

but 12/M= 0.055923

M = 12/0.055923=215 g/mol