How many protons are present in 1.8 g of NH_ | vedclass

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(C) The number of protons in one $NH_{4}^{+}$ ion is calculated as: $7 (	ext{for } N) + 4 (	ext{for } H) = 11$ protons.The molar mass of $NH_{4}^{+}

Vedclass · Accepted Answer

$1.1$

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(C) The number of protons in one $NH_{4}^{+}$ ion is calculated as: $7 (	ext{for } N) + 4 (	ext{for } H) = 11$ protons.The molar mass of $NH_{4}^{+}$ is $14 + 4 = 18 \ g/mol$.The number of moles of $NH_{4}^{+}$ in $1.8 \ g$ is $\frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$.The number of $NH_{4}^{+}$ ions is $0.1 \ N_{A}$.Therefore,the total number of protons is $11 	imes 0.1 \ N_{A} = 1.1 \ N_{A}$.

How many protons are present in $1.8 \ g$ of $NH_{4}^{+}$? (Express in terms of $N_{A}$)

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