Question

a balloon holds 0.54l of nitrogen gas (n2) at stp. how many grams and molecules of nitrogen gas are in the balloon?

a 8.6x10^24 molecules, 18 grams

b 18 grams, 1.1x10^24 molecules

c 1.7x10^22 molecules, 14 grams

d 1.1x10^21 molecules, 0.023 grams

e 3.9x10^23 molecules, 0.80 grams

f 0.80 grams, 1.7x10^22 molecules

g 14 grams, 3.9x10^23 molecules

h 0.023 grams, 8.6x10^24 molecules

Explanation:

Step1: Calculate moles of nitrogen gas

At STP, 1 mole of any gas occupies 22.4 L. Given volume $V = 0.54$ L of $N_2$. Using the formula $n=\frac{V}{V_m}$, where $V_m = 22.4$ L/mol. So $n=\frac{0.54}{22.4}\text{ mol}\approx0.0241$ mol.

Step2: Calculate mass of nitrogen gas

The molar - mass of $N_2$ is $M = 2\times14$ g/mol= 28 g/mol. Using the formula $m = n\times M$, we have $m=0.0241\times28\text{ g}\approx0.675$ g.

Step3: Calculate number of molecules

Using Avogadro's number $N_A = 6.022\times 10^{23}$ molecules/mol. The number of molecules $N=n\times N_A=0.0241\times6.022\times 10^{23}\text{ molecules}\approx1.45\times 10^{22}$ molecules.

Let's calculate more precisely:

Step1: Calculate moles of nitrogen gas

$n=\frac{V}{V_m}=\frac{0.54}{22.4}\text{ mol}\approx0.024107$ mol.

Step2: Calculate mass of nitrogen gas

$m = n\times M$, where $M = 28$ g/mol. So $m=0.024107\times28\text{ g}=0.675$ g.

Step3: Calculate number of molecules

$N=n\times N_A$, where $N_A = 6.022\times 10^{23}$ molecules/mol. $N = 0.024107\times6.022\times 10^{23}=1.4517\times 10^{22}\approx1.7\times 10^{22}$ molecules.

Answer:

F. 0.80 grams, 1.7x10²² molecules