From 161 g of a NO2g sample- 1-8066 - 1024 molecules of NO2g are removed- Find the volume of the left over NO2g at STP-

The number of moles of NO_2(g) nbsp;present in 161 g of NO_2(g), nbsp; n = massmolar mass = 161 g46 g/mol = 3.5 mol The number of moles of NO_2(g) removed ...

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Byju's Answer

Standard XIII

Chemistry

Mole Concept

From 161 g of...

Question

From 161 g of a $N O_{2} (g)$ sample, $1.8066 \times 10^{24}$ molecules of $N O_{2} (g)$ are removed. Find the volume of the left over $N O_{2} (g)$ at STP.

11.2 L

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22.2 L

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22.4 L

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5.6 L

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Solution

The correct option is A 11.2 L
The number of moles of $N O_{2} (g)$ present in 161 g of $N O_{2} (g)$ ,
$n = \frac{m a s s}{m o l a r m a s s} = \frac{161 g}{46 g / m o l} = 3.5 m o l$

The number of moles of $N O_{2} (g)$ removed,
$n = \frac{g i v e n n u m b e r o f m o l e c u l e s}{N_{A}} = \frac{1.8066 \times 10^{24}}{6.022 \times 10^{23}} = 3 m o l e$

The number of moles of $N O_{2} (g)$ left,
= 3.5 - 3 = 0.5 mol

Volume of $N O_{2} (g)$ left at STP.
$V = n \times 22.4 L$
$V = 0.5 \times 22.4 L = 11.2 L$

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From 161 g of a NO2(g) sample, 1.8066×1024 molecules of NO2(g) are removed. Find the volume of the left over NO2(g) at STP.

From 161 g of a $N O_{2} (g)$ sample, $1.8066 \times 10^{24}$ molecules of $N O_{2} (g)$ are removed. Find the volume of the left over $N O_{2} (g)$ at STP.