Posted by **yipez** on Monday, November 15, 2010 at 10:33pm.

Okay, so I got question 1 but I don't know how to solve question 2. Since they are tied in, I'll post both of them.

1. When 0.43g of NaNO2 is allowed to react with an excess of HSO3NH2, the volume of N2 gas produced is 0.145 L. After the reaction is completed, the temperature of water is 24.6 (degrees Celsius) and that of gas is 24.21 (degrees Celsius). The atmospheric pressure is 100.72 kPa. Calculate the molar volume of N2 at 298.15K and 100kPa. You may not use PV=nRT

So for this part of the question, I used P1V1=P2V2.

V2 = 0.146L

That part is okay, but then...

2) When 0.5963g of a sample containing NaNO2 and some inert material is allowed to react with excess HSO3NH2, the volume of N2 gas produced is measured. The volume of N2 gas corrected to 298.15K and 100kPa is calculated to be 0.146L. Find the mass percent of NaNO2 in the sample using the molar volume of N2 at 298.15K and 100kPa (from the previous question).

Please help. :(

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