Each month, science graduates have the opportunity to test their skills and solve exam questions for current science students. These questions appear in the University of Science’s monthly newsletter, Discovery Monthly. His first science question of the month was published in the 2022 alumni magazine Discovery.
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Presented by Vice Dean and Professor of Chemistry Vince Catalano, December’s science question of the month is drawn from the Chemistry 181 exam.
At Méthode Champenoise, grape juice is fermented in wine bottles to make champagne. How many grams of glucose (C6H.12〇6) 500 mL of CO2(g) Produced at 0.862 atm and 310 K?
Fermentation reactions are:
Ha6fire12〇6 (water) →2CH3CH2Oh(aq) +2 CO2(g)
To solve this problem, we need to use the ideal gas law equation:
PV = nRT
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the gas constant, and T is the temperature.
CO pressure is given2 We can assume that the gas is 0.862 atm, has a volume of 500 mL and a temperature of 310 K. I would like to know the number of moles of CO.2 The gas produced can be used to calculate the number of moles of glucose fermented.
First, we need to convert the gas volume from mL to L. This is because the ideal gas law equation requires the volume to be in liters. This can be done by dividing the volume in mL by 1000.
Next, we need to determine the value of the gas constant R. The value of the gas constant depends on the units used for pressure, volume, and temperature. The correct value for the gas constant to use in this problem is 0.0821 L*atm/(mol*K). This value is suitable to use when pressure is in atm, volume is in liters, and temperature is in Kelvin.
Finally, we can plug the values we have into the ideal gas law equation and solve for n, the number of moles of CO.2 Gas Produced:
PV = nRT (0.862 atm)(0.5 L) = n(0.0821 L*atm/(mol*K))(310 K)
n = 0.0169 moles
Now that we know the number of moles of CO2 Using the gas produced, the equilibrium chemical equation for the fermentation reaction, the moles of glucose fermented can be calculated. According to the balanced formula, for every mole of glucose fermented, 2 moles of CO2 produced. So the moles of glucose fermented will be:
0.0169 mol of CO2 / (2 mol CO2/mol glucose) = 0.0085 mol glucose
Finally, we can convert the moles of glucose to grams using the molar mass of glucose which is 180.16 g/mol.
0.0085 mol glucose * 180.16 g/mol = 1.53 g glucose
Therefore, 500 mL of CO2 Gas is produced at 0.862 atm and 310 K during the fermentation process and a total of 1.53 g glucose is fermented.
December’s winner was Daniel Phillips, who submitted the correct answer first. Congrats Daniel!
Phillips, ’02 Ph.D. class (Chemistry) correctly answered the Chemistry 181 exam question.
Phillips grew up in Olinda, California. He returned there in his 2014. Phillips is currently a professor of chemistry at a college in St. Mary, California, where he primarily teaches organic and general chemistry, and occasionally biochemistry. Phillips has authored ten of his research papers, three of which have been published since his college graduation.
“Go pack!” he said.