# A solution contains 3.95g of CS2 and 2.43g CH3COCH3.vapor pressures at 35°C of pure CS2 and pure CH3COCH3 are 616 torr and 332 torr respectively.Assume ideal ,calculate vapor pressure of each component n the total vapor pressure?

##### 1 Answer

The total vapor pressure is

#### Explanation:

This problem requires you to have some idea about **Raoult's Law**, which allows you to calculate the *partial vapor pressure* of a liquid that's part of a mixture by using the vapor pressure of the pure substance and its mole fraction in the mixture.

#color(blue)(P_"A" = chi_"A" * P_"A"^@)" "# , where

*pure* component.

You need to use the molar masses of the two substances to find how many moles of each you have. For carbon disulfide you have

#3.95color(red)(cancel(color(black)("g"))) * "1 mole"/(76.13color(red)(cancel(color(black)("g")))) = "0.0519 moles CS"""_2#

For acetone, you have

#2.43color(red)(cancel(color(black)("g"))) * "1 mole"/(58.08color(red)(cancel(color(black)("g")))) = "0.0418 moles"# #("CH"""_3)_2"CO"#

The *total number of moles* in the mixture is

#n_"total" = n_(CS_2) + n_((CH_3)_2CO)#

#n_"total" = 0.0519 + 0.0418 = "0.0937 moles"#

The *mole fraction* of carbon sulfide is

#chi_(CS_2) = n_(CS_2)/n_"total"#

#chi_(CS_2) = (0.0519color(red)(cancel(color(black)("moles"))))/(0.0937color(red)(cancel(color(black)("moles")))) = "0.554"#

The *mole fraction* of acetone is

#chi_"acetone" = n_"acetone"/n_"total"#

#chi_"acetone" = (0.0418color(red)(cancel(color(black)("moles"))))/(0.0937color(red)(cancel(color(black)("moles")))) = "0.446"#

This means that the partial vapor pressure of carbon disulfide is

#P_(CS_2) = chi_(CS_2) * P_(CS_2)^@#

#P_(CS_2) = 0.554 * "616 torr" = color(green)("341 torr")#

The partial vapor pressure of acetone will be

#P_"acetone" = 0.446 * "332 torr" = color(green)("148 torr")#

The **total vapor pressure** will be

#P_"total" = P_(CS_2) + P_"acetone"#

#P_"total" = "341 torr" + "148 torr" = color(green)("489 torr")#