Question

b. Determine how many ml of MeOH to use, in order to have a 0.3M solution. You may want to review what “0.3M” means. Show your work.

c. Given the amount of camphor ( 200 mg) we are using in this experiment, please determine how many mg of sodium borohydride to use in this reaction. We would like you to use 5.2 molar equivalents of this reagent. This means 5.2 times the mmol of camphor we are using. As an example: for 110.0 mg of camphor,142 mg of NaBH4 would be used (see if you can confirm this result). For complete credit, your work needs to be clearly drawn out!

Answer 1

b) 0.3M solution means 0.3 moles of MeOH present in a litre of solution.

Now, Molar mass of MeOH is 32g therefore mass of MeOH present in solution is 0.3*32=9.6g.

We also know that density of MeOH is 0.8 g/mL and volume= mass/density, hence

Volume of MeOH=9.6/0.8=12 mL

Therefore, 12mL of MeOH must be added to 988mL of water to have 0.3M solution.

c) Molar Mass of Camphor= 152.23g/mol and Molar Mass of NaBH₄=37.83g/mol

verifying the example: No. of mmoles of camphor (n₁)= mass of camphor in mg/ Molar Mass of camphor in g

n₁ =110/152.23=0.72mmoles

and No. of mmoles of NaBH₄ (n₂)=mass of NaBH₄ in mg/ Molar Mass of NaBH₄ in g

n₂ = 142/37.83=3.75mmoles

given, n₂=5.2 n₁ which can be verified from above

Going back to question:

given Mass of Camphor used is 200mg

Therefore, n₁ =200/152.23= 1.314mmoles

Hence n₂= 5.2 n₁ =5.2*1.314=6.83 mmoles

Thus, Mass of NaBH₄ required= 6.83*Molar Mass of NaBH₄= 6.83*37.83= 258.38 mg