What is the molarity of a solution prepared by dissolving 141.6 g of citric acid, #C_3H_5O(COOH)_3#, in water and then diluting the resulting solution to 3500.0 mL?

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Answer 1 · 2016-02-29T13:10:18

.2106#M# #C_3H_5O(COOH)_3#

Molarity is defined as the mols of a compound/element over 1 liter of solution

#M#(Molarity)= #(Mols)/(1L solution)#

you first need to convert 141.6g of citric acid into mols

#C# = 12.01g * 6 = 72.06g
#H# = 1.0079g * 8 = 8.0632g
#O# = 16.00g * 7 = 112.0g
#C_3H_5O(COOH)_3#= 192.12g

141.6g #C_3H_5O(COOH)_3# #(1mol C_3H_5O(COOH)_3)/(192.12g)#=.7370 mol #C_3H_5O(COOH)_3#

We not have the mols of #C_3H_5O(COOH)_3#, and we are given it is diluted to 3500.0 mL

3500.0 mL #(1Liter)/(1000mL)#=3.5000#L#

plug both the mols of #C_3H_5O(COOH)_3# and #L# of #C_3H_5O(COOH)_3# into

#M#(Molarity)= #(Mols)/(1L solution)#

#M#(Molarity)= #(.7370 mol)/(3.5000L solution)#

#M#= .2106

.2106#M# #C_3H_5O(COOH)_3#