A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL stock solution. A 10.00-mL sample of this stock solution is then placed in a 50.00-mL volumetric flask...

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL stock solution. A 10.00-mL sample of this stock solution is then placed in a 50.00-mL volumetric flask and diluted to the mark with water. What is the molality of the new solution?

Expert Answers
gsenviro eNotes educator| Certified Educator

Ammonium sulfate has a chemical formula of `(NH_4)_2 SO_4`  and molecular mass of 132.14 g. The molarity of a solution is the ratio of number of moles of solute to the volume of solution in liters. 

Here, we have 10.8 g of ammonium sulfate and hence the number of moles of solute are:

number of moles = mass / molecular mass = 10.8 g / 132.14 g/mol = 0.082 moles

Volume of solution = 100 mL = 0.1 L

Thus, molarity of ammonium solution = 0.082 moles / 0.1 l = 0.82 M

Now, this solution is diluted by taking 10 ml of this solution and adding 40 ml water to it. We can use the following relationship

C1V1 = C2V2

where C1 and C2 are the concentrations of solutions before and after dilution and V1 and V2 are the volumes of solutions before and after the dilution.

Thus, 0.82 M x 10 ml = C2 x 50 ml

or, C2 = 0.164 M

Thus, the resulting solution has a concentration of 0.164 M.

Hope this helps.

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