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A solution is prepared by mixing 150 g of sodium citrate (FW = 294g/mol) to a...

A solution is prepared by mixing 150 g of sodium citrate (FW = 294g/mol) to a total volume of 500 mL. A second reagent is prepared by using 10 mL of this and making a total volume of 50 mL. What is the molarity and %W/V of each solution.

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Homework Answers

Answer #1

First Solution:-

Mass = 150g

Molar mass = 294g/mol

Moles = mass/molar mass = 150g/294g/mol = 0.510moles

Volume = 500mL = 500/1000L = 0.5L

Molarity = moles/Volume in L = 0.510moles/0.5L = 1.02M

%W/V = mass in grams / volume of solution = 150g/500mL = 0.3 or 30%

Second solution

10mL of first solution is used so,

Molarity = 1.02M

Volume = 10mL = 10/1000 = 0.01L

Moles = Molarity*Volume = 1.02M*0.01L = 0.0102moles

Volume of the solution = 50mL = 50/1000 = 0.05L

Molarity = moles/Volume = 0.0102moles/0.05L = 0.204M

Mass of the solute = moles*molar mass = 0.0102moles*294g/mol = 2.999g

%W/V = 2.999/50 = 0.059 or 5.9%

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