3.42 , g of sucrose (C_{12}H_{22}O_{11}) is d | vedclass

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(A) $1$. Molar mass of sucrose $(C_{12}H_{22}O_{11})$ = $(12 	imes 12) + (22 	imes 1) + (11 	imes 16) = 144 + 22 + 176 = 342 \, g/mol$.$2$. Moles o

Vedclass · Accepted Answer

$6.09 	imes 10^{22}$

Vedclass · Answer

(A) $1$. Molar mass of sucrose $(C_{12}H_{22}O_{11})$ = $(12 	imes 12) + (22 	imes 1) + (11 	imes 16) = 144 + 22 + 176 = 342 \, g/mol$.$2$. Moles of sucrose = $\frac{3.42 \, g}{342 \, g/mol} = 0.01 \, mol$.$3$. Oxygen atoms from sucrose = $0.01 	imes 11 	imes N_A = 0.11 	imes 6.022 	imes 10^{23} = 6.6242 	imes 10^{22}$ atoms.$4$. Moles of water $(H_2O)$ = $\frac{18 \, g}{18 \, g/mol} = 1 \, mol$.$5$. Oxygen atoms from water = $1 	imes 1 	imes N_A = 6.022 	imes 10^{23}$ atoms.$6$. Total oxygen atoms = $(0.066242 	imes 10^{23}) + (6.022 	imes 10^{23}) = 6.088242 	imes 10^{23} \approx 6.09 	imes 10^{23}$ atoms. (Note: The provided option $A$ is $6.09 	imes 10^{22}$,which appears to be a typo in the question's options; however,based on standard calculation,the result is $6.09 	imes 10^{23}$).

$3.42 \, g$ of sucrose $(C_{12}H_{22}O_{11})$ is dissolved in $18 \, g$ of water $(H_2O)$. The number of oxygen atoms in the solution is:

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