Calculate the amount of moles in CH3COOH which is 5%acedic acid and C3H7OH which is 70% usp? This is what I got.....For the amount of 50 ml -> 50mL = 50g ->50g/18.016 = 2.7 moles in 50 ml of H2O

For the amount of 50 ml -> 50mL = 50g ->50g/60.09500 g = .83 moles in 50 ml
Thanks

To calculate the amount of moles in CH3COOH and C3H7OH, you first need to convert the given percentages to grams.

For CH3COOH:
Since it is 5% acetic acid, you can assume that you have 5g acetic acid in every 100g of CH3COOH solution. Therefore, if you have 50 mL (which is equal to 50g) of CH3COOH, you can calculate:

Percentage of acetic acid = 5%
Mass of acetic acid = (Percentage of acetic acid/100) * Mass of CH3COOH
= (5/100) * 50g
= 2.5g

Next, convert the mass of acetic acid to moles:
Molar mass of acetic acid (CH3COOH) = (1*12) + (3*1) + 12 + (16*2) = 60.052 g/mol

Number of moles of acetic acid = Mass of acetic acid / Molar mass of acetic acid
= 2.5g / 60.052 g/mol
= 0.0416 moles

So, in 50 mL of CH3COOH with 5% acetic acid, there are approximately 0.0416 moles of acetic acid.

For C3H7OH:
Similarly, assuming that you have 70g of USP-grade C3H7OH in 100g of C3H7OH solution, you can calculate:

Percentage of USP = 70%
Mass of USP = (Percentage of USP/100) * Mass of C3H7OH
= (70/100) * 50g
= 35g

Convert the mass of USP to moles:
Molar mass of USP (C3H7OH) = (3*12) + (8*1) + 16 = 60.095 g/mol

Number of moles of USP = Mass of USP / Molar mass of USP
= 35g / 60.095 g/mol
= 0.582 moles

So, in 50 mL of USP-grade C3H7OH with 70% USP, there are approximately 0.582 moles of C3H7OH.

It seems like your calculation for CH3COOH has an error in the molar mass (it should be 60.052 g/mol) and the calculated value (0.0416 moles), but your calculation for C3H7OH is correct with 0.582 moles.