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Part A How many atoms of nitrogen are there in a 260.0 mL sample of N, O, gas that has a pressure of 755.0 mmHg at 28.0°C? Ex
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Answer #1

Using the formula PV = nRT

Here, P = 755 mmHg = 0.9934 atm

V = 260 mL = 0.26 L

T = 273 + 28 = 301 K

R = 0.082 L.atm/K.mol

Now using the above relation.

n = (0.9934 x 0.26)/(0.082 x 301) = 0.0105 mol

Now 1 mol of N2O4 contains = 6.022 x 1023 molecules

0.0105 mol of N2O4 contains = 0.0105 x 6.022 x 1023 molecules = 6.322 x 1021 molecules

Now 1 molecule of N2O4 contains number of nitrogen = 2

6.322 x 1021 molecules of N2O4 contains number of nitrogen = 2 x 6.322 x 1021 = 1.2644 x 1022   

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