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QUESTION

# Determine vapor pressure of solution at ##55^@"C"## that contains 34.2g ##NaCl## in 376mL of water? the vapor pressure of pure water at ##55^@"C"## is 118.1 torr.

The vapor pressure of the solution will be 115 torr.

You're dealing with a solution that consists of a non-volatile , sodium chloride, and water, which will be your .

The general idea is that the of a solution that contains a non-volatile solute will be lower than the vapor pressure of th pure solvent, under the same conditions.

This is known as Raoult' Law, which is mathematically expressed as

##P_"solvent" = chi_"solvent" * P_"solvent"^0##, where

##P_"solvent"## - the vapor pressure of the solvent above the solution that contains the non-volatile solute; ##chi_"solvent"## - fraction of the solvent; ##P_"solvent"^0## - the vapor pressure of the pure solvent.

Since you know that the vapor pressure of pure water at ##55^@"C"## is equal to 118.1 torr, you only need to determine the mole fraction of water in the solution.

To do that, you need the value of its at that temperature, which you can find here:

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

So, use water's volume and to calculate its mass

##rho = m/V => m = rho * V##

##m_"water" = 0.9857"g"/cancel("mL") * 376cancel("mL") = "370.62 g"##

Use the molar masses of water and sodium chloride to determine how many moles of each you have

##34.2cancel("g") * "1 mole NaCl"/(58.44cancel("g")) = "0.5852 moles"## ##NaCl##

and

##370.62cancel("g") * "1 mole water"/(18.02cancel("g")) = "20.57 moles"## ##H_2O##

The total number of moles present in the solution will be

##n_"total" = n_(NaCl) + n_(H_2O)##

##n_"total" = 0.5852 + 20.57 = "21.16 moles"##

The mole fraction of water will be

##chi_"water" = n_"water"/n_"total" = (20.57cancel("moles"))/(21.16cancel("moles")) = "0.972"##

The vapor pressure of the solution will thus be

##P_"solvent" = 0.972 * "118.1 torr" = "114.79 torr"##

Rounded to three , the answer will be

##P_"solvent" = color(green)("115 torr")##