Acetone, C3H6O, is the main ingredient of nail polish remover. A solution is made up by adding 35.0mL of acetone(d=0.790g/mL) to 50.0mL of ethyl alcohol,C2H6O(d=0.798 g/mL). Assume that the volumes are additive. From this information calculate the molarity of acetone in the solution.

ok, I got .0046mol/l, is that correct?

No, I don't think so.

grams acetone = volume x density = 35.0 mL x 0.790 g/mL = 27.65 grams.

moles acetone = g/molar mass = 27.65/58.09 = 0.476.

molarity = moles/L = 0.476/0.085 = 5.5998 which rounds to 5.60 M.

To calculate the molarity of acetone in the solution, we need to first find the number of moles of acetone in the solution.

1. Calculate the mass of acetone:
mass = volume × density
mass of acetone = 35.0 mL × 0.790 g/mL = 27.65 g

2. Convert the mass of acetone to moles using its molar mass:
molar mass of acetone = 12.01 g/mol (C) + 3(1.008 g/mol) (H) + 16.00 g/mol (O) = 58.08 g/mol
moles of acetone = mass / molar mass
= 27.65 g / 58.08 g/mol = 0.476 mol

Now that we have the number of moles of acetone in the solution, we can calculate the molarity:

3. Calculate the volume of the solution:
volume of solution = volume of acetone + volume of ethyl alcohol
= 35.0 mL + 50.0 mL = 85.0 mL

4. Convert the volume of the solution to liters:
volume of solution = 85.0 mL × (1 L / 1000 mL) = 0.085 L

5. Calculate the molarity of acetone:
molarity = moles of solute / volume of solution
= 0.476 mol / 0.085 L
= 5.6 mol/L

Therefore, the molarity of acetone in the solution is 5.6 mol/L.

Based on your calculation, you stated that the molarity is 0.0046 mol/L, which is not correct. The correct molarity should be 5.6 mol/L. Please recheck your calculation.