Calculate what volume of your assigned alcohol ( MW = 88.15 g/mol; BP = 118-120C

Question

Calculate what volume of your assigned alcohol (MW = 88.15 g/mol; BP = 118-120C; density = 0.812 g/mL) corresponds to a mass of 15 grams. You will also need to determine the volume of bleach that needs to be used (MW: 74.44 g/mol, density: 1.02 g/mL). If the molarity is 5.25 % wt/vol, (typical bleach sold in stores is 5.25% wt/vol with a molarity of 0.74M), 115 mL is the volume used. For our case, we will be using 8.25% wt/vol. How much volume will be required? A 10% molar excess of bleach (NaOCl) to alcohol is required. Finally, calculate the volume of glacial acetic acid that must be used. For every 0.01 moles of alcohol, 0.5 mL of acid is required. Calculate what volume of your assigned alcohol (MW = 88.15 g/mol; BP = 118-120C; density = 0.812 g/mL) corresponds to a mass of 15 grams. You will also need to determine the volume of bleach that needs to be used (MW: 74.44 g/mol, density: 1.02 g/mL). If the molarity is 5.25 % wt/vol, (typical bleach sold in stores is 5.25% wt/vol with a molarity of 0.74M), 115 mL is the volume used. For our case, we will be using 8.25% wt/vol. How much volume will be required? A 10% molar excess of bleach (NaOCl) to alcohol is required. Finally, calculate the volume of glacial acetic acid that must be used. For every 0.01 moles of alcohol, 0.5 mL of acid is required. Calculate what volume of your assigned alcohol (MW = 88.15 g/mol; BP = 118-120C; density = 0.812 g/mL) corresponds to a mass of 15 grams. You will also need to determine the volume of bleach that needs to be used (MW: 74.44 g/mol, density: 1.02 g/mL). If the molarity is 5.25 % wt/vol, (typical bleach sold in stores is 5.25% wt/vol with a molarity of 0.74M), 115 mL is the volume used. For our case, we will be using 8.25% wt/vol. How much volume will be required? A 10% molar excess of bleach (NaOCl) to alcohol is required. Finally, calculate the volume of glacial acetic acid that must be used. For every 0.01 moles of alcohol, 0.5 mL of acid is required.

Explanation / Answer

Ans. #I. Required volume of alcohol = Required mass / Density

                                                = 15.0 g / (0.812 g/ mL)

                                                = 18.473 mL

#II. Required amount of bleach = 115.0 mL of 5.25 % (w/v)

Available bleach solution = 8.25 % (w/v)

Now, using –

            C1V1 (required bleach) = C2V2 (available bleach)

            Or, 5.25 % x 115.0 mL = 8.25 % x V2

            Or, V2 = (5.25 % x 115.0 mL) / 8.25 %

            Hence, V2 = 73.182 mL

Therefore, required volume of 8.25% bleach = 73.182 mL

#III. Given, a 10% molar excess of bleach to alcohol is required.

Moles of alcohol used = Mass/ Molar mass = 15.0 g / (88.15 g/mol) = 0.170 mol

So,

Required moles of bleach = moles of alcohol + 10% of moles of alcohol

                                    = 0.170 mol + (10.0 % of 0.170 mol)

                                    = 0.187 mol

# Required mass of bleach = required moles x MW

= 0.187 mol x (74.44 g/mol)

= 13.92 g

# Required volume of bleach = Required mass / Available bleach concertation

                                                = 13.92 g / (8.25 %, w/v)

                                                = 13.92 g / (8.25 g / 100 mL)

                                                = 168.72 mL

#IV. Required vol. of acid = (0.5 mL acid / 0.01 mol alcohol) x Moles of alcohol taken

                                                = (0.5 mL acid / 0.01 mol alcohol) x 0.170 mol alcohol

                                                = 8.5 mL

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