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The volume of nitrogen gas Vm (at STP) r...

The volume of nitrogen gas `Vm` (at STP) reqired to cover a sample of silica gel with a monomolecular layer is `129cm^(3)g^(-1)` of gel. Calculate the surface area per gram of the gel if each nitrogen molecule occupies `16.23xx10^(-20)m^(2)` .

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To solve the problem, we will follow these steps: ### Step 1: Calculate the number of nitrogen molecules in 129 cm³ of nitrogen gas at STP. At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22,400 mL. Therefore, we can use the following relationship to find the number of moles in 129 cm³ of nitrogen gas: \[ \text{Number of moles of } N_2 = \frac{\text{Volume of } N_2}{\text{Molar volume at STP}} = \frac{129 \, \text{cm}^3}{22400 \, \text{cm}^3/mol} ...
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The volume of nitrogen gas Um (measured at STP) required to cover a sample of silica get with a mono-molecular layer is 129 cm^(3) g^(-1) of gel. Calculate the surface area per gram of the gel if each nitrogen molecule occupies 16.2 xx 10^(-20) m^(2).

At 1 atm and 273 K the volume of nitrogen gas required to cover a sample of silica gel, assuming Langmuir monolayer adsorption , is found to be 1.30 cm^(3)g^(-1) of the gel. The area occupied by a nitrogen molecule is 0.16nm^(2) . Find out the no. of surface sites occupied per molecule of N_(2) .

Volume of N_(2) at 1 atm, 273 K required to form a monolayer on the surface of iron catalyst is 8.15ml//gm of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies 16 xx 10^(-22)m^(2) ? [Take : N_(A)=6xx10^(23) ]

AT STP the volume of nitrogen gas required to cover a sample of silica gel, assuming Langmuir monolayer adsorption, is found to be 1.3^30^1 of the gel. The area occupied by a nitrogen molecule is 0.1 0^2 What is the surface area per gram of silica gel? [Given N_A =6 diamond ^2 ]

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2.0 g of charcoal is placed in 100 mL of 0.05 M CH_(3)COOH to form an adsorbed mono-acidic layer of acetic acid molecules and thereby the molarity of CH_(3)COOH reduces to 0.49 . The surface area of charcoal is 3xx10^(2) m^(2)g^(-1) . The surface area of charcoal is adsorbed by each molecule of acetic acid is a. 1.0xx10^(-18)m^(2) b. 1.0xx10^(-19)m^(2) c. 1.0xx10^(13)m^(2) d. 1.0xx10^(-22)m

One gram of charcoal adsorbs 400 " mL of " 0.5 M acetic acid to form a mono layer and the molarity of acetic acid reduced to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. The surface area of charcoal is 3.01xx10^(2)m^(2)g^(-1) .

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