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Topic: Finding Unknown Solid's Molar Mass  (Read 11903 times)

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Offline Sis290025

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Finding Unknown Solid's Molar Mass
« on: January 12, 2006, 08:52:19 PM »
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*

Offline sdekivit

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Re:Finding Unknown Solid's Molar Mass
« Reply #1 on: January 13, 2006, 02:25:37 PM »
looks fine to me ;)

Offline plu

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Re:Finding Unknown Solid's Molar Mass
« Reply #2 on: January 13, 2006, 11:14:08 PM »
In my opinion, this question is very icky :P  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  ;)

Offline jdurg

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Re:Finding Unknown Solid's Molar Mass
« Reply #3 on: January 19, 2006, 11:51:19 AM »
In my opinion, this question is very icky :P  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.   ;D
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Offline Borek

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Re:Finding Unknown Solid's Molar Mass
« Reply #4 on: January 19, 2006, 12:40:33 PM »
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.   ;D

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  :P
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