# 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?

Feb 29, 2016

.2106$M$ ${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}$

#### Explanation:

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

$M$(Molarity)= $\frac{M o l s}{1 L s o l u t i o n}$

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
${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}$= 192.12g

141.6g ${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}$ $\frac{1 m o l {C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}}{192.12 g}$=.7370 mol ${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}$

We not have the mols of ${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}$, and we are given it is diluted to 3500.0 mL

3500.0 mL $\frac{1 L i t e r}{1000 m L}$=3.5000$L$

plug both the mols of ${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}$ and $L$ of ${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}$ into

$M$(Molarity)= $\frac{M o l s}{1 L s o l u t i o n}$

$M$(Molarity)= $\frac{.7370 m o l}{3.5000 L s o l u t i o n}$

$M$= .2106

.2106$M$ ${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3}$