Question

A solution of 1.00 g of anhydrous aluminum chloride, $\mathrm{AlCl}_3$, in 50.0 g of water freezes at $-1.11^{\circ} \mathrm{C}$. Does the molar mass determined from this freezing point agree with that calculated from the formula? Why?

   A solution of 1.00 g of anhydrous aluminum chloride, $\mathrm{AlCl}_3$, in 50.0 g of water freezes at $-1.11^{\circ} \mathrm{C}$. Does the molar mass determined from this freezing point agree with that calculated from the formula? Why?
 
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Chemistry
Chemistry
Raymond Chang, Jason… 14th Edition
Chapter 12, Problem 94 ↓
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A solution of 1.00 g of anhydrous aluminum chloride, $\mathrm{AlCl}_3$, in 50.0 g of water freezes at $-1.11^{\circ} \mathrm{C}$. Does the molar mass determined from this freezing point agree with that calculated from the formula? Why?
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Transcript

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00:02 This is question 104 from chapter 13 and we have one gram of in hydrantrous aluminum chloride and we have to calculate the molar mass from the freezing point and compare it to the calculated formula.
00:19 Okay, so the calculated formula molar mass of aluminum chloride is 133 .5 grams per mold.
00:48 We are going to first calculate the molality with what we have.
00:58 So, molality is equal to the molest of solvent over kilograms.
01:11 Okay, so they give us one gram, and we are going to divide that by the molar mass.
01:38 And they give us 50 grams, so to convert that to kilograms, is going to be, 0 .050...
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