If the vapor pressure of an aqueous solution containing 6.00 moles of a nonvolatile solute has a vapor pressure of 19.8 torr, and given that the vapor pressure of water at room temperature is 23.7 torr, how many total moles are present in solution? Your answer should have three significant figures.

36.5 mol

Explanation:

The vapor pressure of a solution of a non volatile solute in  water is given by Raoult´s law:

P H₂O = χ H₂O x P⁰ H₂O

where  χ H₂O  is the mole fraction of water in the solution and P⁰ H₂O

In the turn the mole fraction is given by

χ H₂O = mol H₂O / total # moles = mol H₂O /ntot

Thus

P H₂O = mole H₂O / n tot  x   P⁰ H₂0

now the mol of H₂O is equal n tot - 6  mol solute

Plugging the values given in the question and  solving the resultant equation

19.8 torr = ( ntot - 6 ) x 23.7 torr / n tot

19.8 ntot  = 23.7 ntot - 142.2

ntot = 36.5 ( rounded to 3 significant figures )

oh it is true if thats what you are asking

explanation:

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