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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")