\(50g\) \(of\) \(C_2H_4O_2\) \(in\) \(200g\) \(H_2O\)
so:
\(\frac{50}{200+50}+100\%= \frac{50*100}{250}\%=20\%\)
\(MW of acid = 2*C+4*H+2*O = 2*12+4*1+2*16=\)
\(=24+4+32=60g/mol\)
so:
\(\frac{50g}{60g/mol}\approx0.83moles\)
it means that
in \(50g\) of acid there is \(\approx0.83\)moles of acid
\(MW of H_2O = 2*H+O=2*1+16=2+16=18g/mol\)
so:
\(\frac{200g}{18g/mol}\approx11.11moles\)
it means that:
in \(200g\) of water there is \(\approx11.11\)moles of water
therefore:
\(\frac{0.83mol}{11.11mol+0.83mol}= \frac{0.83mol}{11.94mol}=0.069\)
So your answers are:
\(20\%\)
and the mole fraction is:
\(0.069\)
