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
Sep 29, 2015

Answer:

Composition by mass: #"31.8% Q"# and #"68.2% H"""_2"O"#.

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 #"100900 Nm"""^(-2)#, which means that you can use the vapor pressure of pure water at this temperature to get the vapor pressure of pure Q at this temperature

#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 #M_"M Q"/M_"M water"#, the ratio between the molar mass of #Q# and the molar mass of water, to get

#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 #Q# and the mass of water in the mixture

#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# in the mixture will be

#"% Q" = 100% - "%water" = 100% - 68.2% = color(green)("31.8% Q")#