How many grams the in 4.5times 10^22 molecules of Ba(NO_(2))_(2)

To find the mass of $4.5\times 10^{22}$ molecules of $Ba(NO_{2})_{2}$, we need to follow these steps:<br /><br />1. Calculate the molar mass of $Ba(NO_{2})_{2}$.<br />2. Convert the number of molecules to moles.<br />3. Calculate the mass of the substance.<br /><br />Step 1: Calculate the molar mass of $Ba(NO_{2})_{2}$.<br /><br />The molar mass of $Ba(NO_{2})_{2}$ is calculated by adding the molar masses of its constituent elements:<br /><br />Molar mass of $Ba(NO_{2})_{2}$ = Molar mass of Ba + 2 * Molar mass of N + 4 * Molar mass of O<br /><br />Molar mass of $Ba(NO_{2})_{2}$ = 137.33 g/mol + 2 * 14.01 g/mol + 4 * 16.00 g/mol<br />Molar mass of $Ba(NO_{2})_{2}$ = 261.34 g/mol<br /><br />Step 2: Convert the number of molecules to moles.<br /><br />To convert the number of molecules to moles, we use Avogadro's number, which is $6.022\times 10^{23}$ molecules/mol.<br /><br />Number of moles = Number of molecules / Avogadro's number<br />Number of moles = $4.5\times 10^{22}$ molecules / $6.022\times 10^{23}$ molecules/mol<br />Number of moles ≈ 0.0747 mol<br /><br />Step 3: Calculate the mass of the substance.<br /><br />Mass = Number of moles * Molar mass<br />Mass = 0.0747 mol * 261.34 g/mol<br />Mass ≈ 19.58 g<br /><br />Therefore, the mass of $4.5\times 10^{22}$ molecules of $Ba(NO_{2})_{2}$ is approximately 19.58 grams.