1a) A sucrose solution is prepared to a final concentration of 0.220 M . Convert

Question

1a) A sucrose solution is prepared to a final concentration of 0.220 M . Convert this value into terms of g/L, molality, and mass % (molecular weight, MWsucrose = 342.296 g/mol ; density, soln = 1.02 g/mL ; mass of water, mwat = 944.7 g ). Note that the mass of solute is included in the density of the solution. Express the concentrations in grams per liter, molality, and mass percent to three significant figures separated by commas.

1b)

Suppose 207.9 mg of PbCl2 was added to 15.0 mL of water in a flask, and the solution was allowed to reach equilibrium at 20.0 oC. Some solute remained at the bottom of the flask after equilibrium, and the solution was filtered to collect the remaining PbCl2, which had a mass of 59.4 mg . What is the solubility of PbCl2 (in g/L)?

Express the concentration in grams per liter to three significant figures.

Explanation / Answer

1. Basis : 1 liter of solution.   Given molarity = 0.22 M = 0.22 moles/liter. Hence moles of glucose =0.22

Moles =mass/molar mass, mass of solution = molar mass* moles =0.22*342.296 gm=75.3 gm

Hence concentration in g/L= 75.3g/L

Density of solution = 1.02 g/ml, mass of solution = volume of solution (ml)* density= 1000*1.02= 1020 gm, mass of glucose = 75.3gm, mass of water =1020-75.30=944.7 gm= 944.7/1000kg =0.945

Mass % of glucose in the solution = 100*75.3/1020=7.38%

Molality = moles of solute/kg of solvent = 0.22 moles /0.945 kg=0.233m

2.

Mass of PbCl2 initially= 207.9 mg, mass of PbCl2 after dissolution = 59.4 mg

Mass of PbCl2 dissolved = 207.9-59.4=148.5 mg= 148.5*10-3 gm

This is dissolved in 15ml. 1000ml= 1L, 15ml= 15*10-3 L

Hence solubility = 148.5*10-3 g/(15*10-3L) = 9.9 g/L

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