Question #b4f98

1 Answer
Feb 3, 2016

Answer:

#M_r=55.25#

Explanation:

I'll refer to the acid as #HA#.

Start with the equation:

#HA+NaOHrarrNaA+H_2O#

This tells us that 1 mole of #NaOH# will neutralise 1 mole #HA#.

We can find the number of moles of #NaOH# if we can find its volume.

Density #rho# is given by:

#rho=m/v#

#:.v=m/rho=32.68/0.9705=33.673"ml"#

Since #c=n/v# then we can find the number of moles of #NaOHrArr#

#n_(NaOH)=cxxv=1xx33.673/1000=3.3673xx10^(-2)#

Because the acid and alkali react in a 1:1 ratio we can say that the number of moles of #HA# must be the same.

#:.n_(HA)=3.3673xx10^(-2)#

Now we need to find the mass of #HA# used.

I will use the %w/v value of 62.02 which @Stefan found in another version of the question.

#:.m_(HA)/V_(soln)=0.6202#

You stated that 3ml of solution was used so:

#m_(HA)=0.6202xx3=1.8606"g"#

#:.3.3673xx10^(-2)"mol"# weighs #1.8606"g"#

So 1 mole weighs:

#(1.8606)/(3.3673xx10^(-2))=55.25"g"#

#M_r=55.25#