Question

You must show your work to get full credit for the problem solving questions below. Partial credit will be given for partial work. 1. A 150-ml solution contains 55 g of C,H,OH as a solute. What is the molarity of the solution? C2 points) 2. What is the molality of a solution prepared by dissolving 40.0 g of NaCl in 400 g of water? (2 points) 3. You are asked to prepare a 350 mL standard solution of 0.25M NaOH. How would you make the solution? (2 points) 4. A sample of rubbing alcohol contains 31 g isopropyl alcohol (C,H,OH) and 25 g of H20. What is the mole fraction of H2O? (2 points) 5. Calculate the %(w/w) of a solution prepared by dissolving 10.5 grams of table sugar in 110 mL of water. The density of the water is 1.00 g/mL.

Answer 1

1)

volume = 150 mL = 0.150L

mass of C3H7OH = 55g

molar mass of C3H7OH = 60 g/mol

number of moles = mass / molar mass = 55 / 60 = 0.917 moles

Molarity = number of moles / volume in L = 0.917/0.150 = 6.11 M

Molarity of the solution = 6.11M

2)

mass of solute ( NaCl) = 40.0g

molar mass of NaCl = 58.44 g/mol

number of moles of NaCl = 40.0 / 58.44 = 0.684 moles

mass of solvent ( H2O ) = 400 grams = 0.400 Kg

Molality = number   of moles of solute / mass of solvent in Kg

Molality = 0.684 / 0.400 = 1.71 m

Molality = 1.71 m

3)

volume = 350 mL = 0.350L

Molarity of NaoH = 0.25M

molar mass of NaOH = 40.0 g/mole

Molarity = number of moles / volume in L

number of moles = Molarity x volume in L

number of moles of NaOH = 0.25M x 0.350L = 0.0875 moles

but number o f moles = mass / molar mass

mass of NaOH = number of moles of NaOH x molar mass = 0.0875 x 40.0 = 3.5 grans]

mass of NaOH required for the preparing solution = 3.5 grams

4)

mass of isopropyl alcohol = 31 g

molar mass of isopropyl alcohol = 60.0 g/mole

number of moles of isopropyl alcohol = 31/60.0 = 0.517 moles

mass of water = 25 grams

molar mass of water = 18.0 g/mole

number of moles of water = 25/18.0 = 1.389 moles

Total number of moles = 0.517 + 1.389 = 1.906 moles

Mole fraction of isopropyl alcohol = number of moles of isopropyl alcohol / Total number of moles

Mole fraction of isopropyl alcohol = 0.517 / 1.906 = 0.271

Mole fraction of isopropyl alcohol = 0.271

5)

mass of table sugar = 10.5 g

volume of water = 110 ml

density of water = 1.00 g/ml

mass of 1 ml of water = 1.0 g

mass of 110 ml of water = 100 x 1.0= 110 g

total mass of the solution = 10.5 + 110 = 120.5 g

%(w/w) = mass of solute x100 / total mass of the solution

%(w/w) = 10.5 x 100 / 120.5 = 8.71 %