# Question 98e37

Oct 2, 2016

Let the molar mass of the given mono-basic acid of (empirical formula ${C}_{3} {H}_{4} {O}_{3}$) is $\text{ "m" g/mol}$
If 2.2 g is dissolved in water to produce 1 L solution then the strength of the acid solution will be $= 2.2 \text{g/L} = \frac{2.2}{m} M$

Since the acid given is mono basic one the mole ratio in which the acid and NaOH will react is 1:1.

25 mL 0.5M NaOH $\equiv \frac{25 \times 0.5}{1000} \text{ mol } N a O H = 0.0125 m o l N a O H$

As this 25mL NaOH solution completely neutralizes 1 L acid solution containing $\frac{2.2}{m} \text{ mol}$ acid we can write

$\frac{2.2}{m} : 0.0125 = 1 : 1$

$\implies m = \frac{2.2}{0.0125} = 176 \text{ g/mol}$

The given empirical formula of acid is ${C}_{3} {H}_{4} {O}_{3}$

Let its molecular formula be ${\left({C}_{3} {H}_{4} {O}_{3}\right)}_{n}$

So by this MF its molar mass becomes

=3xx12+4xx1+3xx16)n=88n " g/mol"#

So

$88 n = 176$

$n = \frac{176}{88} = 2$

Hence the Molecular formula of the acid is

${\left({C}_{3} {H}_{4} {O}_{3}\right)}_{2} = {C}_{6} {H}_{8} {O}_{6}$