Question

question 3
how many grams will 5.854e21 formula units of mercury(ii) nitrate weigh?
2.995e - 5 g
1.144e48 g
1.086e43 g
3.156 g

Explanation:

Step1: Calculate the molar mass of mercury(II) nitrate ($Hg(NO_3)_2$)

• Molar mass of $Hg$ = 200.59 g/mol
• Molar mass of $N$ = 14.01 g/mol, and there are 2 $N$ atoms in $Hg(NO_3)_2$, so mass from $N$ = $2\times14.01=28.02$ g/mol
• Molar mass of $O$ = 16.00 g/mol, and there are 6 $O$ atoms in $Hg(NO_3)_2$, so mass from $O$ = $6\times16.00 = 96.00$ g/mol
• Molar mass of $Hg(NO_3)_2=200.59+28.02 + 96.00=324.61$ g/mol

Step2: Use Avogadro's number ($N_A = 6.022\times10^{23}$ formula units/mol) to find the number of moles

Number of moles ($n$) of $Hg(NO_3)_2=\frac{5.854\times 10^{21}\text{ formula units}}{6.022\times 10^{23}\text{ formula units/mol}}=\frac{5.854}{6.022}\times10^{- 2}\text{ mol}\approx0.00972\text{ mol}$

Step3: Calculate the mass ($m$) using the formula $m=n\times M$ (where $M$ is the molar mass)

$m = 0.00972\text{ mol}\times324.61\text{ g/mol}\approx3.156$ g

Answer:

3.156 g