Question

3.) how many formula units (particles of agno₃) are in 5.50 grams of agno₃? (1 point)
8.67×10²¹ formula units
0.0032 formula units
6.14 and 10²² formula units
1.93×10²² formula units

Explanation:

Step1: Calculate molar mass of AgNO₃

The molar mass of Ag (silver) is approximately 107.87 g/mol, N (nitrogen) is 14.01 g/mol, and O (oxygen) is 16.00 g/mol. For AgNO₃, molar mass $M = 107.87+14.01 + 3\times16.00=169.88$ g/mol.

Step2: Calculate number of moles

The number of moles $n$ of AgNO₃ is given by the formula $n=\frac{m}{M}$, where $m = 5.50$ g and $M = 169.88$ g/mol. So $n=\frac{5.50}{169.88}\approx0.0324$ mol.

Step3: Calculate formula - units

We know that 1 mole of any substance contains Avogadro's number ($N_A = 6.022\times 10^{23}$) of formula - units. So the number of formula - units $N=n\times N_A$. Substituting $n = 0.0324$ mol and $N_A=6.022\times 10^{23}$ formula - units/mol, we get $N=0.0324\times6.022\times 10^{23}\approx1.95\times 10^{22}$ formula - units.

Answer:

D. $1.93\times 10^{22}$ formula units (the closest value to our calculated result)