When compound Q with a relative molecular mass of 120 is steam distilled,the mixture boils at 98°C. The vapor pressure of water is 94300Nm^-2 at this temperature. It the atm pressure is 100900Nm^-2,calculate the composition by mass of the distillate?
1 Answer
Composition by mass:
Explanation:
The same idea applies here, as well.
The number of moles each component of the mixture contributes to the total numbe of moles in the vapor will be proportional to their respective partial pressures.
The total pressure at which the mixture boils will be
P_"total" = P_Q + P_"water"
P_Q = "100900 Nm"""^(-2) - "94300 Nm"""^(-2) = "6600 Nm"""^(-2)
This means that you have
n_Q/n_"water" = P_Q^0/P_"water"^0
Since you need to find the composition by mass of the mixture, multiply both sides of the equation by
n_Q/n_"water" * M_"M Q"/M_"M water" = P_Q^0/P_"water"^0 * M_"M Q"/M_"M water"
Since you know that
M_M = m/n implies m = n * M_M , you have
m_Q/m_"water" = P_Q^0/P_"water"^0 * M_"M Q"/M_"M water"
Plug in your values to find the ratio between the mass of
m_Q/m_"water" = (6600color(red)(cancel(color(black)("Nm"""^(-2)))))/(94300color(red)(cancel(color(black)("Nm"""^(-2))))) * (120color(red)(cancel(color(black)("g/mol"))))/(18.02color(red)(cancel(color(black)("g/mol"))))
m_Q/m_"water" = 0.4661
Now, you know that the total mass of the mixture is
m_"total" = m_Q + m_"water"
This means that you can write
m_"total" = overbrace(m_"water" * 0.4661)^(color(blue)(=m_Q)) + m_"water"
m_"total" = 1.4661 * m_"water"
The percent composition of water in the mixture will be
"%water" = m_"water"/m_"total" xx 100
"%water" = color(red)(cancel(color(black)(m_"water")))/(1.4661 * color(red)(cancel(color(black)(m_"water")))) xx 100 = color(green)("68.2 % water")
The percent composition of
"% Q" = 100% - "%water" = 100% - 68.2% = color(green)("31.8% Q")
