Determine the molarity and mole fraction of a 1.15 m solution of acetone (ch3coch3) dissolved in ethanol (c2h5oh). (density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) assume that the final volume equals the sum of the volumes of acetone and ethanol.

The molarity is 0.85M

The mole fraction = 0.05

Explanation:

Step 1: Data given

Molarity of acetone = 1.15 m

Density of acetone = 0.788 g/cm³

Density of ethanol = 0.789 g/cm³

Molar mass of acetone = 58.08 g/mol

Step 2: Calculate number of moles acetone

Molality = moles solute / kg solvent = moles acetone / kg ethanol. (To make it easy, we will suppose we have 1 kg ethanol)

1.15 m = 1.15 moles acetone / 1 kg ethanol

Step 3: Calculate mass of acetone

1.15 moles acetone * (58.08 g/mol) = 66.792 g acetone

Step 4: Calculate volume of acetone

66.792 g acetone / 0.788 g/mL acetone = 84.76 mL acetone

Step 5: Calculate volume of ethanol

1000 g ethanol / 0.789 g/mL ethanol = 1267.43 mL ethanol

Step 6: Calculate total volume solution

Total solution volume = 84.76 + 1267.43 = 1352.2 mL  = 1.3522L

Step 7: Calculate molarity

Molarity of acetone = moles acetone /volume  solution = 1.15moles / 1.3522L

Molarity = 0.85 M

Step 8: Calculate moles ethanol

moles ethanol= mass/ molar mass = 1000g/ 46.0g/mol = 21.74 moles

Step 9: Calculate mole fraction

mole fraction acetone = (moles acetone / total moles) = (1.15 / (1.15 + 21.74)) = 0.05

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates

c, liquids evaporate, creating a gas