\(n_{alanin\left(TT\right)}=0,4\left(mol\right)\\ PT:\left(Ala\right)_5+4H_2O\rightarrow5Ala\\ n_{\left(Ala\right)_n}=n_X=\dfrac{37,3}{373}=0,1\left(mol\right)\\ \Rightarrow n_{Ala\left(LT\right)}=0,1.5=0,5\left(mol\right)\\ \Rightarrow H=\dfrac{0,4}{0,5}.100=80\%\)