Question

unit 3 - the mole quiz
show your all your work. record your final answer on the line provided. include appropriate units and sig figs.

1. determine the molar mass of barium nitrate
2. what is the mass of 4.53 x 10^24 molecules of nacl
3. determine the volume of 15.62g of carbon dioxide at stp.
4. find the number of particles of oxygen gas in 19.4l (at stp)

Explanation:

Step1: Determine the molar - mass of barium nitrate ($Ba(NO_3)_2$)

The molar mass of $Ba = 137.33\ g/mol$, $N = 14.01\ g/mol$, $O = 16.00\ g/mol$.
$M_{Ba(NO_3)_2}=137.33 + 2\times(14.01+3\times16.00)$
$=137.33 + 2\times(14.01 + 48.00)$
$=137.33+2\times62.01$
$=137.33 + 124.02$
$=261.35\ g/mol$

Step2: Calculate the mass of $4.53\times10^{24}$ molecules of $NaCl$

First, use Avogadro's number $N_A = 6.022\times10^{23}\ mol^{-1}$ to find the number of moles.
$n=\frac{N}{N_A}=\frac{4.53\times10^{24}}{6.022\times10^{23}\ mol^{-1}}\approx7.52\ mol$
The molar mass of $NaCl$ is $M_{NaCl}=22.99 + 35.45=58.44\ g/mol$
$m = n\times M=7.52\ mol\times58.44\ g/mol\approx440\ g$

Step3: Determine the volume of $15.62\ g$ of carbon - dioxide at STP

The molar mass of $CO_2$ is $M_{CO_2}=12.01+2\times16.00 = 44.01\ g/mol$
$n=\frac{m}{M}=\frac{15.62\ g}{44.01\ g/mol}\approx0.355\ mol$
At STP, $1\ mol$ of any gas occupies $22.4\ L$.
$V=n\times22.4\ L/mol = 0.355\ mol\times22.4\ L/mol = 7.95\ L$

Step4: Find the number of particles of oxygen gas in $19.4\ L$ at STP

At STP, $1\ mol$ of any gas occupies $22.4\ L$.
$n=\frac{V}{22.4\ L/mol}=\frac{19.4\ L}{22.4\ L/mol}\approx0.866\ mol$
Using Avogadro's number $N = n\times N_A$
$N = 0.866\ mol\times6.022\times10^{23}\ mol^{-1}\approx5.22\times10^{23}$ particles

Answer:

1. $261.35\ g/mol$
2. $440\ g$
3. $7.95\ L$
4. $5.22\times10^{23}$ particles