This is a Walkley-Black organic carbon titration problem, and it is easiest to work through by first finding how much dichromate the soil's organic matter actually used up, then turning that into a carbon percentage.
- Total oxidant added: 10 ml of 1 N potassium dichromate gives 10 milliequivalents of oxidant, added identically to both the blank flask and the soil sample flask.
- What the blank tells us: With no soil present, all 10 meq of dichromate stays unreacted, and it takes 20.1 ml of FAS to titrate it, so 20.1 ml of FAS is equivalent to the full 10 meq dose.
- What the sample titre tells us: The soil sample only needed 17.4 ml of FAS, a drop of 2.7 ml compared with the blank, and this drop exists only because the soil's organic carbon used up some of the dichromate before the back titration began.
- Convert the titre drop to carbon: The fraction of dichromate consumed is 2.7 out of 20.1, about 0.1343, so the soil consumed 0.1343 times 10 meq, which is 1.343 meq of dichromate. Multiplying by 0.003 g of carbon per meq gives 0.00403 g of carbon in the 1 g sample, or 0.403% before correction.
- Apply the recovery factor: The Walkley-Black method only oxidizes a fraction, typically 75 to 77%, of the true organic carbon, so the reading is scaled up by the standard factor of 1.33, giving 0.403 times 1.33, about 0.54%.
The rigorous computed value lands close to 0.54%, and while this does not land exactly on any of the four printed options, it sits nearest to 0.57%, making that the best supported choice, ahead of 0.47%, 0.37% or 0.67%.
Let's summarize:
- The blank titre fixes the true meq value of the dichromate dose.
- The drop from blank to sample titre measures dichromate used by soil carbon, converted using 0.003 g per meq and the 1.33 recovery correction.
So the organic carbon content works out closest to 0.57% among the given options.