Question

directions: read the following lesson from ck12 (making sure to watch the embedded videos), then answer the questions. 10.5 conversion between mass and number of particles 1.) what is the mass of 16.6 x 10^23 molecules of barium nitrate, ba(no3)2? (1 point) 4338 grams 721 grams 2541 grams 927 grams 2.) how many molecules are in 5.00 grams of nh3? (1 point) 6.68 x 10^20 0.294 x 10^23 2.00 x 10^20 1.77 x 10^23

Explanation:

Step1: Calculate molar mass of $Ba(NO_3)_2$

The molar mass of $Ba$ is approximately 137.33 g/mol, $N$ is 14.01 g/mol, and $O$ is 16.00 g/mol. For $Ba(NO_3)_2$, molar mass $M = 137.33+(2\times14.01)+(6\times16.00)=261.35$ g/mol.

Step2: Use Avogadro's number

Avogadro's number $N_A = 6.022\times 10^{23}$ molecules/mol. First find the number of moles $n$ of $Ba(NO_3)_2$. $n=\frac{16.6\times 10^{23}}{6.022\times 10^{23}}$ mol. Then find the mass $m$, $m = n\times M=\frac{16.6\times 10^{23}}{6.022\times 10^{23}}\times261.35\approx721$ g.

Step3: Calculate molar mass of $NH_3$

The molar mass of $N$ is 14.01 g/mol and $H$ is 1.01 g/mol. For $NH_3$, molar mass $M_{NH_3}=14.01 + 3\times1.01=17.04$ g/mol.

Step4: Find moles of $NH_3$

The number of moles of $NH_3$, $n_{NH_3}=\frac{5.00}{17.04}$ mol.

Step5: Find number of molecules of $NH_3$

Using Avogadro's number, the number of molecules $N = n_{NH_3}\times N_A=\frac{5.00}{17.04}\times6.022\times 10^{23}\approx1.77\times 10^{23}$ molecules.

Answer:

1. B. 721 grams
2. D. $1.77\times 10^{23}$