Question

A balloon originally has 0.100 moles of helium and has a volume of 0.500 L. If 0.590 grams of He are added to the balloon, what will the new volume be, in L? (Assume temperature and pressure do not change.)

Answer 1

Answer:

1.24 L

Explanation:

Avogadro's Law

$\frac{V_{1} }{n_{1} } = \frac{V_{2} }{n_{2} }$

Convert 0.590 g of He to moles.

0.590 g x (1 mol/4.00 g) = 0.1475 mol He

Add 0.1475 mol to 0.100 mol since the initial moles (n1) increases by this amount.

0.100 mol + 0.1475 mol = 0.2475 mol He

This will represent the final moles (n2)

The initial volume is given, 0.500 L. Plug in the values.

$\frac{0.500 L}{0.100 mol} = \frac{V_{2} }{0.2465 mol}$

$V_{2} = \frac{(0.500 L)(0.2475 mol)}{(0.100 mol)} \\V_{2} = 1.2375\\V_{2} = 1.24 L$