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

Answer:

See explanation.

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:

$M=\frac{n_{solute}}{V_{solution}}$

Thus:

$n_{solute}=5g*\frac{1mol}{60.052g} =0.083mol\\V_{solution}=V_{solute}+V_{solvent}=5mL+95mL =100mL=0.1L\\M=\frac{0.083mol}{0.1L}=0.83M$

(b) Molality:

$m=\frac{n_{solute}}{m_{solvent}}$

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

$m=\frac{0.083mol}{0.095kg}=0.874m$

(c) Parts by mass:

$\% m/m=\frac{m_{solute}}{m_{solution}}*100\%=\frac{5g}{5g*\frac{1mL}{1.05g} +95g*\frac{1mL}{1g}}*100\% =5.01\%$

(d) Mole fraction:

$x=\frac{n_{solute}}{n_{solution}}$

In this case, the moles of water are required:

$n_{H_2O}=95mL*\frac{1g}{1mL} *\frac{1molH_2O}{18gH_2O}=5.28molH_2O$

Since:

$x=\frac{0.083mol}{0.083mol+5.28mol}=0.155$

Best regards.