Answer:
[tex]m_B=6.55g[/tex]
Explanation:
Hello,
In this case, considering the mole fraction of carbon tetrachloride, we can write:
[tex]0.824=\frac{n_{CCl_4}}{n_{CCl_4}+n_{B}}[/tex]
Whereas the moles of carbon tetrachloride are computed from the given 60.5 g (molar mass 153.82 g/mol):
[tex]n_{CCl_4}=60.5g*\frac{1mol}{153.82g}=0.393mol[/tex]
So we can compute the moles of benzene from the definition of molar fraction:
[tex]n_B=\frac{0.393-0.824*0.393}{0.824} =0.0839mol[/tex]
Now, with the molar mass of benzene (78.11 g/mol) we compute the mass of benzene:
[tex]m_B=0.0839mol*\frac{78.11g}{1mol} \\\\m_B=6.55g[/tex]
Regards.
