<p>Hello, I am completely stuck on this. I need help as I have a test tomorrow and I need to do some practice questions. please help me out!</p>
<p>Answer the following concerning propanone (acetone, C3H6O) a substance used to remove nail polish.</p>
<p>A) Is propanone expected to be soluble in water? Explain.</p>
<p>B) A solution is prepared by combining 30.0 ml propanone (density 0.792 g/ml) with 50.0 ml 2-propanol (C3H7OH, density 0.785 g/ml).
Assume that the volumes are additive.</p>
<p>1) Calculate the percent by mass of 2-propanol in the solution</p>
<p>2) Calculate the percent by volume of propanone in the solution</p>
<p>3) Calculate the mole fraction of propanone in the solution.</p>
<p>D) Compare the expected freezing point of solution described in part (A) to the freezing point of pure 2-propanol. Is the freezing point of the solution expected to be higher than, equal to, or lower than the freezing point of the pure solvent?
EXPLAIN your answer!</p>
<p>A) For a substance to be soluble, the forces between the solute particles must be of similar type/strength as the forces between the solvent particles. Water has van der Waals forces and (more importantly) hydrogen bonding, so only solutes that have hydrogen bonding or are very polar (or ionic) are soluble in water. Propanone is fairly polar, so you would expect it to be soluble.
B)(1) Mass percent A = (mass A)/(total mass solution) x 100%. Use (density x volume) for each substance to find their masses.
B)(2) Volume percent A = (volume A)/(total volume solution) x 100%. 30/80 x 100%.
B)(3) Mole fraction A = mol A/total moles. Use the grams you calculated in B1 and convert to moles, then plug the numbers in.
D) When solvents contain dissolved substances, boiling points are elevated and melting points are depressed. The freezing point will be lower than the pure solvent. (Colligative properties - delta(MP) = m x k x i.)</p>