Question

A sample containing 1.20 moles of Ne gas has an initial volume of 5.50 L. What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change? A) A leak allows one-half of the Ne atoms to escape. Express your answer with the appropriate units. B) A sample of 3.80 moles of Ne is added to the 1.20 moles of Ne gas in the container. Express your answer with the appropriate units. C) A sample of 25.0 g of Ne is added to the 1.20 moles of Ne gas in the container. Express your answer with the appropriate units.

Answer #1

_{1} = 1.20 mol ; then n_{2} = 1.2/2 =
0.6 mol

at n_{1} = 1.20 mol V_{1} = 5.50L so to find volume
at 0.6 mol use the formula below

V_{1} / n_{1} = V_{2} / n_{2}

5.50 / 1.2 = V_{2} / 0.6

2.75 L = V_{2}

B) After addition of Ne = 1.2 mol + 3.8 mol = 5 mol

use the same formula

5.50 / 1.2 = V_{2} / 5

V_{2} = 22.9167 L

C) mol of Ne added = 25 gm / 20.1797 gm/mol = 1.2388 mol

[Molar mass of Ne = 20.1797 gm/mol]

After addition mol of Ne = 1.2 mol + 1.2388 mol = 2.4388 mol

5.50 / 1.2 = V_{2} / 2.4388

V_{2} = 11.1778 L

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