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Standard X

Chemistry

Volume of Gases and Number of Moles

Volume of N...

Question

Volume of $N_{2}$ at NTP required to form a mono layer on the surface of iron catalyst is $8.15 m l / g r a m$ of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies $16 \times 10^{- 22} m^{2}$ .

16 \times 10^{- 22} c m^{2}

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3.5 \times 10^{- 4} c m^{2}

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39 m^{2} / g

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22400 c m^{2}

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Solution

The correct option is D $3.5 \times 10^{- 4} c m^{2}$
$8.15 m l N_{2} = 8.15 \times 10^{- 3} l$ of $N_{2}$
$∴$ moles of $N_{2} = \frac{8.15 \times 10^{- 3}}{22.4} = 0.3638 \times 10^{- 3}$
$∴$ molecules $= 0.364 \times 10^{- 3} \times 6.022 \times 10^{23} = 2.19 \times 10^{20}$
$∴$ surface area covered $= (16 \times 10^{- 29} m^{2}) (2.19 \times 10^{20})$
$= 35.072 \times 10^{- 9} m^{2} = 3.5 \times 10^{- 8} m^{2} = 3.5 \times 10^{- 4} {c m}^{2}$

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Volume of N2 at NTP required to form a mono layer on the surface of iron catalyst is 8.15ml/gram of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies 16×10−22m2.

The correct option is D 3.5×10−4cm28.15mlN2=8.15×10−3l of N2∴ moles of N2=8.15×10−322.4=0.3638×10−3∴ molecules=0.364×10−3×6.022×1023=2.19×1020∴ surface area covered =(16×10−29m2)(2.19×1020)=35.072×10−9m2=3.5×10−8m2=3.5×10−4cm2

Volume of $N_{2}$ at NTP required to form a mono layer on the surface of iron catalyst is $8.15 m l / g r a m$ of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies $16 \times 10^{- 22} m^{2}$ .