Question

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 hour 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 \(\underline{\hspace{2cm}}\).

• A. 15540 mg/L
• B. 8620 mg/L
• C. 6920 mg/L
• D. 1700 mg/L

Hint:

To calculate the concentration of volatile solids, subtract the final mass (after heating) from the initial mass, then divide by the volume of the sample in liters and convert the result to mg/L.

Answer 1

The Correct Option is C

To determine the concentration of total volatile solids (TVS), we need to calculate the mass of volatile solids and then express this mass in mg/L, considering the initial volume of the sample. Step 1: Calculate the mass of total volatile solids.
The total volatile solids are the difference in mass before and after the sample is heated in the muffle furnace. The mass of the sample before heating includes both volatile and fixed solids, while the mass after heating represents only the fixed solids. - Initial weight of the crucible with the wet sample = 55.129 g
- Final weight of the crucible with residue = 54.783 g
- Mass of volatile solids = Initial weight - Final weight \[ \text{Mass of volatile solids} = 55.129 \, \text{g} - 54.783 \, \text{g} = 0.346 \, \text{g} \] Step 2: Convert mass to mg.
Since the mass of volatile solids is in grams, we need to convert it to milligrams: \[ 0.346 \, \text{g} = 346 \, \text{mg} \] Step 3: Calculate the concentration of volatile solids.
The volume of the sample is given as 50 mL. To express the concentration in mg/L, we first convert the volume to liters: \[ 50 \, \text{mL} = 0.050 \, \text{L} \] Now, we can calculate the concentration: \[ \text{Concentration of total volatile solids} = \frac{346 \, \text{mg}}{0.050 \, \text{L}} = 6920 \, \text{mg/L} \] Thus, the correct answer is option (C). Final Answer: 6920 mg/L