P is the atmospheric pressure in the lab which is just 1(atm) V is the volume in
ID: 904566 • Letter: P
Question
P is the atmospheric pressure in the lab which is just 1(atm) V is the volume in liters (divide the ml by 1000) T is the temperature in Kelvin (equal to degree Celsius +273) R is the gas constant 0.082L-atm/mole-kelvin< N is the number of moles of Co2. RistiLs Table Initial Gas Volume Sugar Glucose Fructose Maltose Maltotriose Final Gas Volume 5 minutes (mL) 4.6ml 1.4ml 5.3ml 1.0ml Volume of CO2 produced Final - Initial (mL) minutes (mL) 10 sec=0.1 ml 18 sec-0.1 ml 10 sec-0.1ml 33 sec-0.1ml 2. For each of the sugars fermented by yeast, fll in the chart below to determine the mg of sugar consumed per minute during fermentation. o For column one use to calculate the moles of CO2 produced o Use moles of sugar consumed-moles of CO2 produced ÷ (2 x number of simple sugars in that sugar) to calculate the moles of sugar consumed o Use mg of sugar per minute = (moles sugar)-(MW g/mole)s (1000 mg/g) ÷ (5 minutes) to calculate the mg of sugar fermented per minute Calculations Table Moles of Co2 produced M Moles of Sugar Consumed Mg of sugarímin Sugar Glucose Fructose Maltose Maltotriose to yeast grown commerciall considering that maximizing efficiency is the goal of the industry? 1.Explanation / Answer
Glucose :-
Volume of CO2 produced = 4.6 - 0.1 = 4.5 ml = 0.0045 litres
Now, moles of CO2 produced = (P*V)/(R*T) = (1*0.0045)/(0.0821*298) = 1.84*10-4 moles
Now, C6H12O6 -------> 2CO2 + 2C2H5OH
Moles of glucose consumed = (1/2)*moles of CO2 formed = 9.2*10-5
Molar mass of glucose = 180 g/mole = 180000 mg/mole
Thus, mass of glucose consumed = moles*molar mass = 16.55 mg
2) Fructose :-
Volume of CO2 produced = 1.4 - 0.1 = 1.3 ml = 0.0013 litres
Now, moles of CO2 produced = (P*V)/(R*T) = (1*0.0013)/(0.0821*298) = 5.31*10-5 moles
Now, C6H12O6 -------> 2CO2 + 2C2H5OH
Moles of fructose consumed = (1/2)*moles of CO2 formed = 2.65*10-5
Molar mass of fructose = 180 g/mole = 180000 mg/mole
Thus, mass of fructose consumed = moles*molar mass = 4.782 mg
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