A certain substance X has a normal freezing point of -4.9 degrees celsius and a molal freezing point depression Kf = 6.99 degrees celsius * kg * mol^-1. A solution is prepared by dissolving some urea ((NH_2)_2CO) in 400 g of X. This solution freezes at -7.9 degrees celsius. Calculate the mass of urea that was dissolved.

Question

cuanchi · Answer

ΔTF = KF · b · i, where: ΔTF, the freezing-point depression, is defined as TF (pure solvent) − TF (solution). KF, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. (Note: When conducting experiments, a higher KF value makes it easier to observe larger drops in the freezing point. For water, KF = 1.853 K·kg/mol.[6]) b is the molality (moles solute per kilogram of solvent) i is the van 't Hoff factor (number of ion particles per individual molecule of solute, e.g. i = 2 for NaCl, 3 for BaCl2). https://en.wikipedia.org/wiki/Freezing-point_depression#Calculation in your case i=1 urea is covalent (molecular) compound KF is given in the problem b the molality is defined has moles of solute by kg of solvent. It can be calculated by rearrange the formula. Finally knowing the molality (b) and the grams of solvent given in the problem, you multiply this two values and get the moles of urea in the solution. Then you multiply this by (60 g/mol) the molecular mass of urea and you will find the grams of urea present int he solution. That is the answer to the problem