Question

What mass of CH3COOH is present in a 250 mL cup of 1.25 mol/L solution of vinegar?

Answer 1

If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x

⇒ x =
`(250 ml * 1.25mol)/(1000 ml)`

= 0.3125 mol

∴ moles of CH₃COOH in 250ml is 0.3125 mol

Now, Mass = mole × molar mass

= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol

= 18.75 g

∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is 18.75 g