Question

Please be extra clear on the molarity and molality questions. I’m having a hard time figuring those two out. Thank you :)

1) An aqueous solution was prepared by dissolving 87.750 g of K2CO3 (potassium carbonate) in 225.00 g of water. It was found that a 10.00-ml sample of this solution had a mass of 12.755 g. (FW of K CO. = 138.204) NOTE: You may often see your instructor use the abbreviation FW, which means Formula Weight. Technically, FW is in atomic mass units (amu), and it represents the mass of single formula unit of the compound. For molecular substances, you might see an equivalent abbreviation, MW, meaning Molecular Weight Converting either of these to molar mass is a matter of simply taking the number given and adding g/mol on the end as units. For example, if the FW of K20,- 138.204, then its molar mass - 138204 g/mol. Module 2 - Lecture a) Calculate the density of this solution in g/mL. (2) b) Calculate the mass % of K2CO3 in this solution. (4) c) Calculate the molarity of K2CO3 in this solution. (4) d) Calculate the molality of K2CO3 in this solution. (4)

Answer 1

(a)

~(mi Given that: Mass of K₂CO₃ = 87.750 g Mass of of Water = 225 25.00g (m₂) As lomb of the sample had a mass of mass. a of 12.7554 Density of the solution = Mass volume 12-7559 10 ml 11 1- 275 g/ml

(b)

Mass of the solution = m, + m2 (solute + solvent) 87.750 + 225.00 312.750 g xloo Mass of of K₂CO₃ = Mass of K2CO3 Total Mass of the solution 28.057 % 312-750 750g 87.750 g x

(c)

(6) Molarity of a solution is defined as number of moles of fue solute dissolved in a litre of solution Volume of the Solution Mass of solution Density Ve_312.750 mL 245.30 mL 1.275 0.245 L

Now, According to the Definition of Molarity. M = n = no. of moles volume No. of Moles of K Coz Mars Molecular Mass 87.750 g 138.204 g/mol [co Molar mass - 138,204 N 0-635 mol Molnity CM (M) = 0.635 moli- 0-245 = 2.59 moll

(d)

dll Molality is defined as the No. of moles of so hite dissolved in per kg of solvent molality (m)2 n [m - 225g or 0.225 ig] ma mal by 2.82 mol 0-635 0.225 11 moling

(e)

le Mole fraction (u) No. of Moles of component Total No of roles so.635 225 g 12.5 mol • Moles of K₂CO3 (n) Moles of Water (n) 18 g/mol Total no. of Moles (np) = 0.635 + 12.5 = 13.135.

Mole Fraction of K₂CO₃ (34) = ni -0.635 13- 135 HT 0.0483 Mole Fraction of Water (uz) п, hy 22.5 130135 = 0.9516