By using formula,

${\gamma }_{\mathrm{mixture}}=\frac{\left(\frac{{\mu }_{1}C{p}_1+{\mu }_{2}C{p}_2}{{\mu }_1+{\mu }_2}\right)}{\left(\frac{{\mu }_{1}C{u}_1+{\mu }_{2}C{v}_2}{{\mu }_1+{\mu }_2}\right)}$

${\gamma }_{\mathrm{mixture}}=\frac{{\mu }_{1}C{p}_1+{\mu }_{2}C{p}_2}{{\mu }_{1}C{v}_1+{\mu }_{2}C{v}_2}$

$=\frac{{\mu }_1\left(\frac{{\gamma }_1}{{\gamma }_1-1}\right)R+{\mu }_2\left(\frac{{\gamma }_2}{{\gamma }_2-1}\right)R}{{\mu }_1\left(\frac{R}{{\gamma }_1-1}\right)+{\mu }_2\left(\frac{R}{{\gamma }_2-1}\right)}$

or ${\gamma }_{\mathrm{mixture}}=\frac{{\mu }_1{\gamma }_1\left({\gamma }_2-1\right)+{\mu }_2{\gamma }_2\left({\gamma }_1-1\right)}{{\mu }_1\left({\gamma }_2-1\right)+{\mu }_2\left({\gamma }_1-1\right)}$

It is given, ${\gamma }_{\mathrm{mixture}}=\frac{19}{13}$

${\gamma }_1=\frac{5}{3},{\gamma }_2=\frac{7}{5}$

$n_1=1,n_2=n$

Substituting values, we have 

$\frac{19}{13}=\frac{\frac{5}{3}\left(\frac{7}{5}-1\right)+n\frac{7}{5}\left(\frac{5}{3}-1\right)}{\left(\frac{7}{5}-1\right)+n\left(\frac{5}{3}-1\right)}$

$\Rightarrow \frac{19}{13}=\frac{\frac{5}{3}\times \frac{2}{5}+n\frac{7}{5}\times \frac{2}{3}}{\frac{2}{5}+n\frac{2}{3}}$

$\Rightarrow 19\left(\frac{2}{5}+n\frac{2}{3}\right)=13\left(\frac{2}{3}+n\frac{14}{15}\right)$

$\Rightarrow n=2$

$\therefore$ Number of moles of gas $B=2$.