Answer to Question #185039 in General Chemistry for Jenifer

Question #185039

Solve in complete solution and Explain briefly:

An organic compound is known to be non-volatile and non-electrolyte. A 0.5-g sample is dissolved in water and diluted to 160 mL. The osmotic pressure is measured as 0.05 atm at 30°C. The approximate mass number for this compound is 1400 g/mol.

From the data provided in this problem and knowing that the density of the solution is 1.00 g/mL.


1. Calculate the freezing point of the solutions.

2. Determine if the freezing point change would be a good way to determine the molecular mass of the compound.

3. Would the boiling point change be a better determining factor than the freezing point change?


1
Expert's answer
2021-04-30T02:58:43-0400

"\\Pi = icRT"


"i = \u03a0\/cRT\\\\\n\nT = 30\u00b0C = 273 + 30\u00b0C = 303K\\\\\n\n\u03a0 = 0.05atm = 0.05 \u00d7 101325 = 5066.25"

number of moles "(n) = 0.5\/1400 ="

"0.000357"


concentration (c) = "\\dfrac{0.000357}{160} \u00d7 1000"

"= 0.00223M = 2.23 mol\/m\u00b3"



"i= \\dfrac{5066.25}{ 2.23\u00d78.314\u00d7303}\n=\n0.9"




1. "\u2206T = iK_fm"


"K_f= 1.86K\u00b7kg\/mol"

"i =0.9"

molality (m) = "\\dfrac{0.000357}{0.160}= 0.00223"


"\u2206T = 0.9 \u00d7 1.86\u00d7 0.00223="

"0.00373K"


2. Yes


3. No, because the magnitude of the freezing point depression is larger than the boiling point elevation for the same solvent and the same concentration of a solute.


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