To determine the molar mass of a monocarboxylic acid $A$. $500\;mL$ of an aqueous solution is prepared with $m=6.5\;g\;A$ .
$20\;mL$ of this solution consume in a titration exactly $10\;ml\;NaOH\;0.1\;M$ .
Calculate the molar mass of the acid.
At the equivalence point:
$n_{NaOH}$ $=$ $n_{A}$
$n_{A}$ $=$ $0.020\cdot 0.1$ $=$ $0.0020\; mol $
In $500\; mL$. we have therefore:
$n$ $=$ $\frac{0.0020\cdot 500}{20}$ $=$ $0.050 mol$
$M$ $=$ $\frac{m}{n}$ =
$130\frac{g}{mol}$
