A bit of logic and a bit of chemistry solves the puzzle. The chemical knowledge

Question

A bit of logic and a bit of chemistry solves the puzzle. The chemical knowledge required is that of the general chemistry course. Question Six chemists were seated around a table. Each was a specialist in a major field of chemistry; Bill, Richard, Barbara. Alice, James, and Tom were each (not respectively and not necessarily seated in the listed order), an organic chemist, an analytical chemist, an inorganic chemist, a physical chemist, a biochemist, and a polymer chemist Each chemist had in front of him or her a sealed jar of a chemical whose molecular weight was identical to that chemist's age. Identify the name, field of specialty, and age of each chemist. Clues: 1. The analytical chemist noted that the chemical in front of Richard had a rate of effusion twice as slow as methane. 2. The biochemist, a woman, had a compound whose empirical and molecular formula were identical and was composed of 81.82% C and 18.18% H by weight. 3. The woman sitting on Bill's right had a liquid whose vapor pressure was lowered 4% when 0.1 mol of sugar was dissolved into 84 g of the liquid. 4. Tom, the polymer chemist, was 5 years older than Alice. When 4 g of Tom's chemical (nonelectrolyte) was dissolved in 20 g of H_2 O, the solution boiled at 102.6 degreeC (K_bp for H_2O = 0.52). 5. The inorganic chemist, James, noted that Bill was not a physical chemist, and Bill was 2 years older than the analytical chemist; 0.2368 g of Bill's chemical (a solid monoprotic acid) was able to neutralize 8 mL of 0.40 M Ca(OH)j. 6 James, the youngest, had a 2-carbon hydrocarbon. When a portion of it was burned in oxygen, 52.8 g of COt and 10.8 g of HtO were produced.

Explanation / Answer

Question 3

We use Raoult's formula here:

Pvap = xAP0vap

Where xA is the molar mass of the substance, let's solve for xA first:

xA = Pvap / P0vap

If the vapor pressure lowered 4% this means Pvap = 0.96P0vap

xA = 0.96

Now:

xA = moles of liquid / (moles of sugar + moles of liquid)

0.96 = grams/MW / (0.1 + grams/MW)

Let's solve for MW:

0.96 = (84/MW) / (0.1 + 84/MW)

0.96 ( 0.1 + 84/MW) = 84/MW

0.096 + 80.64/MW = 84/MW

0.096 = 84/MW - 80.64/MW

0.096 = (1/MW)(84-80.64)

0.096/3.36 = 1/MW

MW = 35 g/mol

Then, this scientist is 35 years old.

Question 4

This is a problem for increasing the boiling point of a substance, we follow the formula:

b = kbpm

Where b is the change in boiling point of the solvent, in this case 2.6°C (because the boiling point of water is 100°C) and m = molality = moles of solute / kg of solvent.

b = kbp (moles/kg solvent)

Let's remember that moles = grams/MW;

b = (kbp*grams / MW*kg solvent)

Let's solve for MW:

MW = (kbp*grams /b*kg solvent)

MW = (0.52*4 / 2.6*0.02)

MW = 40 g/mol

Then, Tom is 40 years old.

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