Ans: 64*10^(-4) ltr/hr Q 15 The respiration of the suspension of yeast cells was measured by observing the - Chemistry - States of Matter

Question

Ans: 64*10^(-4) ltr/hr

Q 15 The respiration of the suspension of yeast cells was measured
by observing the decrease in pressure of gas above the cell
suspension The apparatus show that the gas was confined to a
constant volume, 16 ml and the entire pressure change
was caused by uptake of O2 by the cells The pressure was
measured in a manometer, the fluid of which had a density of I 034
gm/ml The entire apparatus was immersed in a thermostat at 37°C
In a 30 min observation period the fluid in open side of manometer
dropped 37 mm neglecting the solubility of O2 in yeast
suspension, compute rate of O2 consumed by the cells in
It of O2 (S T P ) per hour

Vartika Jain · Accepted Answer

Dear Student,

First we need to convert the observed pressure drop to mmHg     =(37 mm)(1
034 g/cm313 6 g/cm3) = 2
8 mmHg in 30 minSo, in an hour it will be 5
6 mmHgWith 5
6 mmHg as the oxygen partial pressure, change it to STPV2=P1V1T2T1P2     =(5
6 mmHg)(16 cm3)(273 K)(310 K)(760 mmHg)     = 0
10 cm3     =1
0×10-4 dm3 =1 0×10-4 L