Explore BrainMass

Titration, solution content and molarity

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

An experiment calls for using 20 mL of toluene. (density=0.8668 g/mL, molecular weight=92.139 g/mol). What mass of toluene, and how many moles are involved?

An experiment calls for 50.0 of benzoyl chloride to be used (density=1.212 g/mL). What volume of benzoyl chloride is needed?

An aqueous solution of nitric acid has a density of 1.41 g/mL and contains 70.0% HNO3 by weight (wt/wt). What is the molarity of this solution?

https://brainmass.com/chemistry/compounds/titration-solution-content-molarity-588859

SOLUTION This solution is FREE courtesy of BrainMass!

1. Volume, V = 20 mL
density, d = 0.8668 g/mL

Hence, mass of toluene, m = volume * density = V*d = 20*0.8668 = 17.336 g

molecular weight of toluene, M = 92.139 g/mol

Hence, number of moles involved in reaction = mass/molecular wt = m/M = 17.336/92.139 = 0.188 mol

2. NOTE: I am assuming 50.0 gm of benzoyl chloride

mass of benzoyl chloride, m = 50.0 g

density of benzoyl chloride, d = 1.212 g/mL

Volume of benzoyl chloride, V = mass/density = m/d = 50.0/1.212 = 41.254 mL

3. Density of HNO3 solution, d = 1.41 g/mL
i.e., mass of 1 mL of solution = 1.41 g

As HNO3 is 70% (wt/wt)

1.41 g of solution, contains HNO3 = 1.41*0.7 = 0.987 g HNO3

i.e., 1 mL solution contains HNO3 = 0.987 g

Hence,
1 L solution contains HNO3 = 1000*0.987 = 987 g of HNO3

molar mass of HNO3 = 1 + 14 + 16*3 = 63

Hence, 987 g HNO3 == 987/63 = 15.67 mol

Hence, 1 L of solution contains 15.67 mol of HNO3

Therefore, molarity of the solution = 15.67

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!