A solution of 0.720 g of an organic compound in 100.g of because has a freezing

Question

A solution of 0.720 g of an organic compound in 100.g of because has a freezing point of 5.149 degree C. What are the molality of the solution and the molar mass of the solute? The normal freezing point of benzene is 5.500 degree C and the K_t for benzene is 5.12 degree C/m. molality = _____m molar mass = ______ g/mol Enter your answer in the provided box. Calculate the molar concentration of oxygen in water at 25 degree C for a partial pressure of 0.27 atm. The Henry's law constant for oxygen is 1.3 times 106-1 mol/L middot atm, Report your answer to two significant figures. 0.00035 M The density of an aqueous solution containing 15.0 percent ethanol (C_2H_5OH) by mass is 0.978 g/mL. Calculate the molality of this solution. _____ m Calculate the solution's molarity. _____ m What volume of the solution would contain 0.246 mole of ethanol? ____ mL Calculate the vapor pressure of a solution made by dissolving _84.8 g of area (molar mass = 60.06 g/mol) in 208.5 ml of water at 35 degree C. 0.891 mm Hz What is the magnitude of vapor-pressure lowering? ____ mmHg

Explanation / Answer

1)
Given:
delta Tf = (5.500 - 5.149) oC = 0.351 oC
Kf =5.12 oC/m
mass of solvent in Kg = 0.1 Kg
mass of solute = 0.72 g

use:
delta Tf = Kf*mb
0.351 = 5.12mb
mb= 0.069
Answer: molality = 0.069 molal

Now use:
molality = mass of solute/(molar mass of solute * mass of solvent in Kg)
0.069 = 0.72/(MM*0.1)
MM =105.0 g/mol
Answer: molar mass = 105 g/mol

I am allowed to answer only 1 question at a time. So answering 1st one (the one at top left corner)

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