Question

Question #98e37

Answer 1

Let the molar mass of the given mono-basic acid of (empirical formula `C_3H_4O_3`) is `" "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"g/L"=2.2/mM`

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(25xx0.5)/1000" mol "NaOH=0.0125molNaOH`

As this 25mL NaOH solution completely neutralizes 1 L acid solution containing `2.2/m" mol"` acid we can write

`2.2/m:0.0125=1:1`

`=>m=2.2/0.0125=176" g/mol"`

The given empirical formula of acid is `C_3H_4O_3`

Let its molecular formula be `(C_3H_4O_3)_n`

So by this MF its molar mass becomes

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

So

`88n=176`

`n=176/88=2`

Hence the Molecular formula of the acid is

`(C_3H_4O_3)_2=C_6H_8O_6`