A 2.950×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.7 mL . The density of water at 20.0∘C is 0.9982 g/mL. Calculate: a. molality
b. mole fraction of glycerol in this solution
c. the concentration of the glycerol solution in percent by mass
d. the concentration of the glycerol solution in parts per million

a. 2.959x10⁻²m

b. 5.327x10⁻⁴

c. 0.272%

d. 2718 PPM

Explanation:

A solution of 2.950x10⁻²M contains 2.950x10⁻² moles of Glycerol per Liter of solution. As the volume of the solution made was 1.000L, moles of glycerol are 2.950x10⁻².

a. molality: Molality is defined as the ratio between moles of solute (2.950x10⁻²) in kg of solvent. As there are 998.7mL of solvent and density is 0.9982g/mL, kg are:

998.7mL ₓ (0.9982g/mL) ₓ (1kg / 1000g) = 0.9969kg of solvent.

Molality: 2.950x10⁻² moles / 0.9969kg of solvent = 2.959x10⁻²m

b. Mole fraction is the ratio between moles of solute and total moles. Moles of water are:

998.7mL ₓ (0.9982g/mL) ₓ (1mol / 18.01g) = 55.35 moles of water.

Mole fraction glycerol:

2.950x10⁻² moles / (2.950x10⁻²moles + 55.35) = 5.327x10⁻⁴

c. Percent by mass Is the ratio by mass of solute and solution multiplied 100 times.

Mass of glycerol (Molar mass:  92.09g/mol):

2.950x10⁻² moles × (92.09g / mol) = 2.717g of glycerol

Mass of water:

998.7mL ₓ (0.9982g/mL) = 996.9g of water.

Percent by mass:

2.717g of glycerol / (996.9g of water + 2.717g) × 100 = 0.272%

d. Parts per million are defined as the ratio between mg of solute and kg of solution.

mg of 2.717g of glycerol are 2717mg

kg of solution are (996.9g + 2.717g) / 1000 = 0.9996kg

Parts per million:

2717mg / 0.9996kg = 2718 PPM

law of conservation of matter

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong