Question

a bottle of commercial vinegar contains 5% acetic acid, ch3cooh, by volume (95% water). the density of acetic acid is 1.05 g/ml and water is 1.00 g/ml. from this data calculate the concentration of acetic acid in vinegar in: molality, molarity, parts by mass, and the mole fraction.

Answer 1

See explanation.

Step-by-step explanation:

Hello,

For this exercise, it is convenient to assume a basis of 5 g of acetic acid which is the solute and 95 g of water which is the solvent, in such a way, one proceeds as follows:

(a) Molarity:

`M=(n_(solute))/(V_(solution))`

Thus:

`n_(solute)=5g*(1mol)/(60.052g) =0.083mol\\V_(solution)=V_(solute)+V_(solvent)=5mL+95mL =100mL=0.1L\\M=(0.083mol)/(0.1L)=0.83M`

(b) Molality:

`m=(n_(solute))/(m_(solvent))`

In this case, the moles were already computed and the mass of the solvent is in kilograms, thus:

`m=(0.083mol)/(0.095kg)=0.874m`

(c) Parts by mass:

`\% m/m=(m_(solute))/(m_(solution))*100\%=(5g)/(5g*(1mL)/(1.05g) +95g*(1mL)/(1g))*100\% =5.01\%`

(d) Mole fraction:

`x=(n_(solute))/(n_(solution))`

In this case, the moles of water are required:

`n_(H_2O)=95mL*(1g)/(1mL) *(1molH_2O)/(18gH_2O)=5.28molH_2O`

Since:

`x=(0.083mol)/(0.083mol+5.28mol)=0.155`

Best regards.