In Dumas method for estimation of nitrogen, 0.30g of an organic compound gave 50 mL of nitrogen collected at 300 K and mm pressure. Calculate the percentage composition of nitrogen in the compound (Aqueous tension at 300 K is 15 mm).

In Dumas method for estimation of nitrogen, 0.30g of an organic compou

1.	In Dumas method for estimation of nitrogen, 0.30g of an organic compound gave 50 mL of nitrogen collected at 300 K and mm pressure. Calculate the percentage composition of nitrogen in the compound (Aqueous tension at 300 K is 15 mm).
Answer» Solution :Here, mass of the SUBSTANCE taken = 0.30g Volume of NITROGEN collected = 50 ML, Atmospheric pressure = 715mm Hg Room temperature = 300K Aqueous tension at 300 K = 15mm `:.` Actual pressure of the gas (dry gas) = `715 - 15 = 700mm Hg` Step 1. To convert the volume at experimental conditions to volume at STP. `{:("Experimental values",,"At STP"),(P_(1)=700 mm,,P_(2) = 760 mm),(V_(1) = 50 mL,,V_(2) = ?),(T_(1) = 300 K,,T_(2) = 273 K):}` Substituting these values in the gas equation. `(P_(2)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`, we get `(700 mm xx 50 mL)/(300 K) = (760 mm xx V_(2) mL)/(273 K)` or `V_(2) = (273 xx 700 xx 50)/(300 xx 760) = 41.9 mL`. Step 2.To convert the volume at STP into mass. According to the definition of GMV, 22400 mL of nitrogen at STP weigh = 28 g `:. 41.9 mL` of nitrogen at STP will weigh `= (28 xx 41.9)/(22400) g` Step 3. To calculate the percentage of nitrogen. Percentage of nitrogen `= ("Mass of "N_(2) " at STP")/("Mass of the substance taken") xx 100 = (28 xx 41.9 xx 100)/(22400 xx 0.3) = 17.46`