What is the number of molecules in 2.125 text | vedclass

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(C) $1$. Calculate the molar mass of ammonia $(NH_3)$: $M = 14 + (3 	imes 1) = 17 	ext{ g/mol}$.$2$. Calculate the number of moles $(n)$: $n = \frac

Vedclass · Accepted Answer

$7.51 	imes 10^{22}$

Vedclass · Answer

(C) $1$. Calculate the molar mass of ammonia $(NH_3)$: $M = 14 + (3 	imes 1) = 17 	ext{ g/mol}$.$2$. Calculate the number of moles $(n)$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{2.125 	ext{ g}}{17 	ext{ g/mol}} = 0.125 	ext{ mol}$.$3$. Calculate the number of molecules: $	ext{Number of molecules} = n 	imes N_A = 0.125 	imes 6.022 	imes 10^{23} = 7.5275 	imes 10^{22} \approx 7.53 	imes 10^{22}$.Given the options, the closest value is $7.51 	imes 10^{22}$.

What is the number of molecules in $2.125 \text{ g}$ of ammonia $(NH_3)$?

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