3.42 , g of sucrose are dissolved in 18 , g o | vedclass

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(A) The formula for the number of moles is: $	ext{moles} = \frac{	ext{mass}}{	ext{molar mass}}$.The molar mass of sucrose $(C_{12}H_{22}O_{11})$ is

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$6.68 	imes 10^{23}$

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(A) The formula for the number of moles is: $	ext{moles} = \frac{	ext{mass}}{	ext{molar mass}}$.The molar mass of sucrose $(C_{12}H_{22}O_{11})$ is $342 \, g/mol$.The number of moles of sucrose $= \frac{3.42 \, g}{342 \, g/mol} = 0.01 \, mol$.Each molecule of sucrose contains $11$ oxygen atoms.Therefore,the number of oxygen atoms from sucrose $= 0.01 	imes 11 	imes N_A = 0.11 \, N_A$.The molar mass of water $(H_2O)$ is $18 \, g/mol$.The number of moles of water $= \frac{18 \, g}{18 \, g/mol} = 1 \, mol$.Each molecule of water contains $1$ oxygen atom.Therefore,the number of oxygen atoms from water $= 1 	imes 1 	imes N_A = 1 \, N_A$.Total number of oxygen atoms $= 0.11 \, N_A + 1 \, N_A = 1.11 \, N_A$.Using $N_A = 6.022 	imes 10^{23}$,the total number of oxygen atoms $= 1.11 	imes 6.022 	imes 10^{23} = 6.68442 	imes 10^{23} \approx 6.68 	imes 10^{23}$.

$3.42 \, g$ of sucrose are dissolved in $18 \, g$ of water in a beaker. The number of oxygen atoms in the solution are

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