# 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

#### 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 *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

#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")#