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 =  $\frac{250 ml * 1.25mol}{1000 ml}$
        
        =  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