If you wanted to make .5 L of a 1 mole/L (M) of MgSO4 solution, how many grams of MgSO4 would you use?
1 answer:
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Answer:
14.9802 grams of estrogen must be added to 216.7 grams of benzene.
Explanation:
The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.
Where:
= Vapor pressure of pure solvent
= Vapor pressure of the solution
= Number of moles of solvent
= Number of moles of solute
Mass of 0.05499 moles of estrogen :
= 0.05499 mol × 272.4 g/mol = 14.9802 g
14.9802 grams of estrogen must be added to 216.7 grams of benzene.