Question #66c6e

The vapor pressure of the solution will be 115 torr.

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

The general idea is that the vapor pressure 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"` - the mole 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 density at that temperature, which you can find here:

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

So, use water's volume and density 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 sig figs, the answer will be

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