Questions for Exercise 1.4 Qi. A chemist gathers aliquots of water, methanol and
ID: 574007 • Letter: Q
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Questions for Exercise 1.4 Qi. A chemist gathers aliquots of water, methanol and acetic acid; measures the volume and the mass of each aliquot; and then pours the three aliquots together to form a mixture. The chemist then measures the mass and volume of the newly formed mixture. Finally, they calculate the amount of molecules brought into the mixture by each component as well as the total amount of molecules in the mixture. These results are summarized in the table below: Amount Mass Volume (Moles) (g) (mL) Before Mixing water 0.110 2.00 2.00 methanol 0.062 2.00 2.53 acetic acid 0.017 1.00 1.05 Totals Mixture 0.19 5.00 5.40 After Mixing (solution) (a) Based on the chemist's results, determine the relative abundance of acetic acid as a: mass fraction, mole fraction, volume fraction, volume concentration, mass concentration, molarity, and molality Note 1: (The tear sheet from the previous exercise (RO) provides definitions that may be helpful.) Note 2: Help for the molality calculation: Generally the solvent is the most abundant liquid. However in this problem water has more moles, and methanol has a bigger volume. Thus in calculating molality you need to choose the solvent. You can select water, methanol or water + methanol as your solvent. Because the calculated molality will change depending upon what you chose as the solvent, please state your solvent choice along with your answer.Explanation / Answer
Mass fraction of acetic acid in the mixture =mass of acetic acid/total mass= 1/5=0.2
Mole fraction of acetic acid = moles of acetic acid/total moles =0.017/0.19=0.0894
Volume fraction = volume of acetic acid/total volume= 1.05/5.40= 0.194
Volume concentration = volume/volume= 1.05/5.4= 0.194
Mass concentration = mass/volume = 1g/5.4ml = 0.185 g/ml
Molarity = moles of acetic acid/1 liter of solution = 0.017/5.4/1000 =3.15M
Molality= moles of acetic acid/kg of solvent
Solvent considered is a mixture of methanol and water. Mass of methanol and water= 2+2=4 gm
Mass of solvent in Kg= 4/1000 =0.004 kg
Molality= 0.017/0.004 =4.25m
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