Covalent bonds and it is between the hydrogens (hydrogen bonds)
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.
<h3>
Answer: A) 3.5 mol/L</h3>
Explanation:
To determine the molarity, we have to find the number of moles in the volume given, and then extrapolate to find the number of moles that would be in 1 L.
<u>Determine the moles in the given volume</u>
moles of LiCl = mass ÷ molar mass
= 139.9 g ÷ 42.39 g/mol
= 3.30 mol
<u>Find the moles in 1 L</u>
Since 930 mL of LiCl = 3.30 mol
then 1000 mL of LiCl = (3.30 mol × 1000 mL/L) ÷ 930 mL
= 3.55 mol/L
