The number of N atoms in 681 g of C_{7}H_{5} | vedclass

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(B) The molar mass of $C_{7}H_{5}N_{3}O_{6}$ is $(7 	imes 12) + (5 	imes 1) + (3 	imes 14) + (6 	imes 16) = 84 + 5 + 42 + 96 = 227 \ g/mol$.The nu

Vedclass · Accepted Answer

$5418$

Vedclass · Answer

(B) The molar mass of $C_{7}H_{5}N_{3}O_{6}$ is $(7 	imes 12) + (5 	imes 1) + (3 	imes 14) + (6 	imes 16) = 84 + 5 + 42 + 96 = 227 \ g/mol$.The number of moles of $C_{7}H_{5}N_{3}O_{6}$ is $n = \frac{681 \ g}{227 \ g/mol} = 3 \ mol$.Since each molecule of $C_{7}H_{5}N_{3}O_{6}$ contains $3$ atoms of $N$,the total moles of $N$ atoms is $n_{N} = 3 	imes 3 = 9 \ mol$.The number of $N$ atoms is $9 	imes N_{A} = 9 	imes 6.02 	imes 10^{23} = 54.18 	imes 10^{23} = 5418 	imes 10^{21}$.Therefore,the value of $x$ is $5418$.

The number of $N$ atoms in $681 \ g$ of $C_{7}H_{5}N_{3}O_{6}$ is $x \times 10^{21}$. The value of $x$ is $.....$ $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$ (Nearest Integer)

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