If 13.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
1 answer:
6.349 g mass of anhydrous magnesium sulfate will remain.
<h3>What are moles?</h3>A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Molar mass MgSO₄.7 H₂O = 246.52 g/mol
0.0527 moles
Molar mass MgSO₄ = 120.4 g/mol
Mass of anhydrous magnesium sulfate :
( 0.0527 x 120.4 ) => 6.349 g
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Answer:
A, C and D are correct.
Explanation:
Hello.
In this case, since the relationship between the vapor pressure of a solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvent as stated by the Raoult's law:
Since the solute is not volatile, the mole fraction of the solute is not taken into account for vapor pressure of the solution, therefore A is correct whereas B is incorrect.
Moreover, since the higher the vapor pressure, the weaker the intermolecular forces due to the fact that less more molecules are like to change from liquid to vapor and therefore more energy is required for such change, we can evidence that both C and D are correct.
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