Hello,

I just wanted to check with you if what I calculated is right.

I am asked to add 10% sodium hydroxide molar relative to acetaldehyde with a 5 mol / L sodium hydroxide solution.

I know that we put 7 L of acetaldehyde (100%).

1 - Calculating how many mol of acetaldehyde has been put:

data on acetaldehyde:

volume (V): 7 L

density (d): 780 g / L

Molar mass (M): 44.05 g / mol

calculation of the mass in 7 L

d = m * L

m = 780 * 7 g

m = 5460 g

calculation of the number of moles in 7 L

n = m / M

n = 5460 / 44.05

n = 123.95 mol

2 - Calculating the volume of sodium hydroxide that must be added

20% of 123.93 mol => 24.7 mol.

The concentration of sodium hydroxide solution being 5 mol / L, it is necessary to add 24.7 / 5 => 4.95 L of this sodium hydroxide solution to have 20 mol% relative to acetaldehyde.

Are my calculations correct?

Thank you !