# Chemistry- please check my work

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Compute the vapor pressure of water above a solution formed by dissolving 25.00 grams of magnesium sulfate, MgSO4 (MM=120.3) in 95.00 grams of water at 338K. The vapor pressure of pure water at 338K is 188.0 torr.

I took 95.00g H20 divided by 18.0g. which gave me moles for top. I than took those moles and added them to 25.0 divided by 120.3. I than took 5.277 mol H20 divided by 5.4856total moles time 188.0 torr. which i than got 180.88 vapor pressure of water.

• Chemistry- please check my work -

•Find the moles of MgSO4 by dividing 95.00g by the formula mass of MgSO4.
•Multiply the moles of MgSO4 by 3 ions / mole to get the moles of ions in solution.
•Convert 95g of H2O to moles.
•Add moles of ions to moles of H2O to get total moles.
•The moles fraction of H2O is:
X(H2O) = (moles H2O)/(total moles)
•P(solution) = X(H2O)P(H2O)
•P(solution) = X(H2O)(188 mm) = ?

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