Problem Statement:
A 50 mL sample of industrial wastewater is taken into a silica crucible. The empty weight of the crucible is 54.352 g. The crucible with the sample is dried in a hot air oven at 104 °C till a constant weight of 55.129 g. Thereafter, the crucible with the dried sample is fired at 600 °C for 1 h in a muffle furnace, and the weight of the crucible along with residue is determined as 54.783 g. The concentration of total volatile solids is
Solution: (C) 6920
Also Check: Solution of GATE 2021, CE, Set1 (Q.1 to Q.100)
Weight of the volatile solids (Wsolids) = Change in the weight of the sample when dried at 600˚C for 1 hour in muffle furnace
hence, Wsolids = 55.129 – 54.783 = 0.346 gm = 346 mg
The Concentration of volatile solids =Wsolids/Vsample = 346/50 = 6.92 mg/ml = 6920 mg/L
Solution for Q.27 <— —-> Solution Q.29
