molality question?

Question

I need to calculate the molality of HCl based on: a formula weight of 36.465 amu, a density of solution as 1.19 g/mL, a weight of 37.2%, and a molarity of 12.1.

not sure where to go first with this...any help is appreciated. thanks in advance:)

Answer 1

Molality is moles solute /kg solvent, so we need to find out how much of each we have in the solution.

We know we have 12.1 moles HCl / L of solution so we just need to find how many kg of solvent are in 1 L (1000mL) of solution

37.2% (w/w) is 37.2g HCl per 100 g solution =>
You have 100-37.2 =62.8 g solvent per 100 g solution

d=mass/V => V=mass/d= 100/1.19=84.03 mL

So 84.03 mL of solution contain 62.8g solvent
so 1000 mL .. .. .. .. .. .. .. .. .. . .. x

x =(62.8/84.03)*1000=747.35 g solvent

So 1000 mL solution contain 747.35 g solvent
But also 1000 mL solution contain 12.1 mole HCl
So 12.1 mole HCl are in 747.35 g solvent
and m =(12.1/747.35)*1000 =16.19

Answer 2

molality = mole of solute/ kg of solvent?? hmm..

molarity = 12.1 moles/ 1L = 12.1 moles per 1 Liter,
1000ml x 1.19g/ml = 1190gram of solvent.. = 1.19kg

weight of 37.2% of What???
anyway, 37.2% of this Solution is weighed, .372 x1.19kg = .4426kg
12.1 moles / .4426 = 27.34 molality!!!

If the percentage was what it meant, then it should be 27.34 molality , check it... hope it works for u..

Answer 3

molality is moles solute per kg os solutios. you have the first get the moles by using the molarity (moles per litter) and the density. Then get the kg the the percent is what w/w or w/v?