Gen. Chem Kaplan Question

[jdpaul14]

Hey Guys,

I don't understand the solution to this problem:

Q: How many grams of Al2(SO4)3 are needed to make 87.5g of 0.3m Al2(SO4)3 solution? (Atomic weights of Al = 27, S = 32, O = 16)

A:
Formula weight of Aluminum Sulfate = 342 g/mol
Thus, deduce that 102.6 g/mol are required per kilogram of solvent
Total mass of solution produced by dissolving 102.6 g of aluminum sulfate in one kilogram (1000g) of solvent will then be 1102.6 g/mol

102.6 g/1102.6 g = X/87.5g

X = (102.6g*87.5g)/1102.6g


Thanks a lot

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[jdpaul14]

Ok, I figured it out from a previous post, here is the answer...thanks anyway

.3molal x 1kg solvent = mol Al2(SO4)3
.3 mol = X g / MW g/mol ---> .3mol = Xg / 342g ---> X g Al2SO43 = .3 x 342 = 102.6 g Al2(SO4)3 (in a .3 molal solution)
So in .3molal solution there is a total of 1000g(solvent) + 102.6g (solute)= 1102.6 g
Given 87.5 g solution, how much solute (Al2SO43) are there?
87.5g solution/ 1102.6 g solution = Y g Al2SO43 / 102.6 g Al2SO43
Y g = 87.5 x 102.6 / 1102.6