One gram of activated charcoal has a surface area of 1000m2. If complete coverage is assumed, Approximately how much ammonia (in cm3 at NTP) could be adsorbed on the surface of 25g of the charcoal ? (Given : Diameter of NH3 molecule = 0.3nm)

Total surface area available = =(25×103)m2=2.5×108cm2Surface area covered by one molecule of NH3=πr2=[227×(0.32×10−7)2]cm2=(227×0.15×10−14×0.15)cm2=0.0707×1014cm2=7.07×10−16cm2No. of ammonia molecules =2.5×1087.07×10−16=3.536×1023No. of moles of NH3=3.536×10236.023×1023=0.587Volume of NH3 at NTP =(0.587×22400)= =13.282.4cm3

One gram of activated charcoal has a surface area of 1000m2. If complete coverage is assumed, Approximately how much ammonia (in cm3 at NTP) could be adsorbed on the surface of 25g of the charcoal ? (Given : Diameter of NH3 molecule = 0.3nm)

Total surface area available = =(25×103)m2=2.5×108cm2Surface area covered by one molecule of NH3=πr2=[227×(0.32×10−7)2]cm2=(227×0.15×10−14×0.15)cm2=0.0707×1014cm2=7.07×10−16cm2No. of ammonia molecules =2.5×1087.07×10−16=3.536×1023No. of moles of NH3=3.536×10236.023×1023=0.587Volume of NH3 at NTP =(0.587×22400)= =13.282.4cm3

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One gram of activated charcoal has a surface area of 1000m². If complete coverage is assumed, Approximately how much ammonia (in cm³ at NTP) could be adsorbed on the surface of 25g of the charcoal ? (Given : Diameter of NH₃ molecule = 0.3nm)

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10148.8 cm3

11148.8 cm3

12148.8 cm3

13282.4cm3

answer is D.

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Detailed Solution

Total surface area available = $= (25 \times 10^{3}) m^{2} = 2.5 \times 10^{8} c m^{2}$
Surface area covered by one molecule of $N H_{3} = π r^{2}$
$= [\frac{22}{7} \times {(\frac{0.3}{2} \times 10^{- 7})}^{2}] c m^{2} = (\frac{22}{7} \times 0.15 \times 10^{- 14} \times 0.15) c m^{2}$ $= 0.0707 \times 10^{14} c m^{2} = 7.07 \times 10^{- 16} c m^{2}$

No. of ammonia molecules $= \frac{2.5 \times 10^{8}}{7.07 \times 10^{- 16}} = 3.536 \times 10^{23}$
No. of moles of $N H_{3} = \frac{3.536 \times 10^{23}}{6.023 \times 10^{23}} = 0.587$
Volume of $N H_{3}$ at NTP $= (0.587 \times 22400) =$ =13.282.4cm³