Question

how many molecules of nh₄br are in 2.991 g of nh₄br? show the conversions required to solve this problem and calculate the molecules of nh₄br.

Explanation:

Step1: Calculate molar mass of $NH_4Br$

The molar mass of $N$ is $14.01\ g/mol$, $H$ is $1.01\ g/mol$, and $Br$ is $79.90\ g/mol$. For $NH_4Br$, molar mass $M=(14.01 + 4\times1.01+79.90)\ g/mol=97.95\ g/mol$.

Step2: Calculate moles of $NH_4Br$

Use the formula $n=\frac{m}{M}$, where $m = 2.991\ g$ and $M = 97.95\ g/mol$. So $n=\frac{2.991\ g}{97.95\ g/mol}=0.03054\ mol$.

Step3: Calculate number of molecules

Use Avogadro's number $N_A = 6.022\times 10^{23}\ molecules/mol$. The number of molecules $N=n\times N_A$. So $N = 0.03054\ mol\times6.022\times 10^{23}\ molecules/mol=1.81\times 10^{22}$ molecules.

Answer:

$1.81\times 10^{22}$ molecules