An aqueous solution containing 17.5 g of an unknown molecular compound in 100.g water was found to have a freezing point of-1.8°C.

Question

An aqueous solution containing 17.5 g of an unknown molecular compound dissolved in 100 g of water was found to have a freezing point of -1.8°C. Calculate the molar mass of this unknown compound by following these steps. Please show your work.

What is the freezing point change (ΔT) from pure water? Given that the freezing point of water is 0.0°C, calculate ΔT.
What is the freezing point depression constant (Kf) for water?
What is the molality of the solution? Use the formula: 
   [ m = \frac{	ext{moles of solute}}{	ext{kg of solvent}} ]
Solve for the moles of solute.
Knowing the mass of the compound and the moles of the compound, calculate the molar mass of the compound. Use the formula:
   [ 	ext{Molar mass} = \frac{	ext{mass of compound}}{	ext{moles of solute}} ]

Anonymous · Accepted Answer

The answer to your question is 178.6 gExplanation:DataΔT = 1.8 °C mass = 17.5 gmass of water = 100 gKc = 1.86Process1.- Calculate the molality using the following formula     ΔTc = mKcsolve for m     m = ΔTc/Kcsubstitution      m = 1.8/1.86result      m = 0.9682.- Calculate the number of moles     m = # of moles/kg of solventkg of solvent = 0.1 kg      # of moles = m x kg of solvent      # of moles = 0.968 x 0.1      # of moles = 0.09683.- Calculate the molar mass         x g molar mass 1 mol        17.5 g                   0.0968 moles        x = (1 x 17.5)/0.0968       x = 178.6 g

Anonymous · Answer

I’m distilled water all of the dissolved substances mixed in water have been removed by evaporation.

Anonymous

An aqueous solution containing 17.5 g of an unknown molecular compound in 100.g water was found to have a freezing point of-1.8°C.

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