[Sis290025]

12.00 g of an unknown solid is dissolved in 200.0 g benzene. The mixture freezes at 3.45 Celsius. Find the molar mass of the compound. Benzene's freezing point is 5.49 Celsius.


delta T_f = K_f * m

m = mol solute (unknown) / kg solvent (benzene)

m = delta T_f / K_f = (5.49 C - 3.45 C) / (4.90 C/m) = 0.41633 m

mol solute = m * (kg solvent) = 0.41633 m * 0.200 kg = 0.083266 mol

molar mass = g solute/mol solute = 12.00 g / 0.083266 mol = 144 g/mol???

Thanks for any *delete me*

[sdekivit]

looks fine to me

[plu]

In my opinion, this question is very icky  The molar mass of the compound could easily be two, three, or even four times that of the value you calculated depending on the physical properties of the compound itself.  Freezing point depression and boiling point elevation is tricky in this respect as these two physical phenomena depend on the total concentration of solutes in solution, not total number of moles of substances dissolved in solution (a subtle, but nonetheless important distinction).  For example, 1 mole of NaCl in H₂O would dissolve to give not 1 but 2 moles of solute! (1 mole of Na⁺ ions and 1 mole of Cl^- ions)  This would thus depress the freezing point of the solution by twice the amount calculated if ionization were ignored.  However, in your case Sis290025, I believe it would be safe to ignore these thoughts  

[jdurg]

In my opinion, this question is very icky  The molar mass of the compound could easily be two, three, or even four times that of the value you calculated depending on the physical properties of the compound itself.  Freezing point depression and boiling point elevation is tricky in this respect as these two physical phenomena depend on the total concentration of solutes in solution, not total number of moles of substances dissolved in solution (a subtle, but nonetheless important distinction).  For example, 1 mole of NaCl in H₂O would dissolve to give not 1 but 2 moles of solute! (1 mole of Na⁺ ions and 1 mole of Cl^- ions)  This would thus depress the freezing point of the solution by twice the amount calculated if ionization were ignored.  However, in your case Sis290025, I believe it would be safe to ignore these thoughts  

Yup, because benzene is completely non-polar so no ionization at all will take place.  There will be no separations going on so one mole in equals one mole dissolved.  

[Borek]

Yup, because benzene is completely non-polar so no ionization at all will take place.  There will be no separations going on so one mole in equals one mole dissolved.  

In non-polar solvent you may expect exactly the opposite - molar mass determined can be twice higher then in reality. So one mole in equals half mole dissolved