How would your calculated value for the molecular weight of your unknown be affe

Question

How would your calculated value for the molecular weight of your unknown be affected (high, low, no effect) if a small quantity of the unknown stuck to the metal stirring rod and failed to dissolve in the benzophenone? Clearly explain your reasoning. When determining the melting point of your pure benzophenone, some of the molten material solidified high on the thermometer and wire stirrer. How would your calculated value for the molecular weight of your unknown been affected (high, low, no effect) if this benzophenone failed to melt and was not a part of the final mixture in the tube? Clearly explain your reasoning.

Explanation / Answer

Freezing point depression equation

TF = KF · m

Where m is the molality = No. of moles of solute / mass of solution

No. of moles = mass of solute /molar mass

Molar mass = mass of solute / No. of moles

while finding molecular mass

if small amount of solute didn't dissolve in solution it will decrease the No. of moles of solute and hence the molality.

in calculation of Molar mass is inversly proportional to no. of moles

hence it will increase the Molar mass

Hence molar mass will be higher than actual molar mass Answer

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