Question #98e37

1 Answer
P dilip_k
Oct 2, 2016

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#

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