select the correct mass for each of the samples described below. 3.01×10²³ molecules h₂o 8.32×10²⁰ formula…

Explanation:

Step1: Recall Avogadro's number

Avogadro's number ($N_A$) is $6.022\times 10^{23}$ particles/mol.

Step2: Calculate moles for $H_2O$

The number of moles of $H_2O$, $n_{H_2O}=\frac{3.01\times 10^{23}}{6.022\times 10^{23}\text{ mol}^{-1}} = 0.5\text{ mol}$. The molar - mass of $H_2O$ is $M_{H_2O}=(2\times1 + 16)\text{ g/mol}=18\text{ g/mol}$. The mass of $H_2O$, $m_{H_2O}=n_{H_2O}\times M_{H_2O}=0.5\text{ mol}\times18\text{ g/mol}=9\text{ g}$.

Step3: Calculate moles for $CaBr_2$

The number of moles of $CaBr_2$, $n_{CaBr_2}=\frac{8.32\times 10^{20}}{6.022\times 10^{23}\text{ mol}^{-1}}\approx0.00138\text{ mol}$. The molar - mass of $CaBr_2$ is $M_{CaBr_2}=(40 + 2\times79.9)\text{ g/mol}=199.8\text{ g/mol}$. The mass of $CaBr_2$, $m_{CaBr_2}=n_{CaBr_2}\times M_{CaBr_2}=0.00138\text{ mol}\times199.8\text{ g/mol}\approx0.276\text{ g}$.

Step4: Calculate moles for $XeF_6$

The number of moles of $XeF_6$, $n_{XeF_6}=\frac{1.93\times 10^{26}}{6.022\times 10^{23}\text{ mol}^{-1}} = 320.5\text{ mol}$. The molar - mass of $XeF_6$ is $M_{XeF_6}=(131.3+6\times19)\text{ g/mol}=245.3\text{ g/mol}$. The mass of $XeF_6$, $m_{XeF_6}=n_{XeF_6}\times M_{XeF_6}=320.5\text{ mol}\times245.3\text{ g/mol}\approx78628\text{ g}$.

Answer:

For $3.01\times 10^{23}$ molecules $H_2O$: $9\text{ g}$ For $8.32\times 10^{20}$ formula units $CaBr_2$: $0.276\text{ g}$ For $1.93\times 10^{26}$ molecules $XeF_6$: $78628\text{ g}$