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CBSE Questions for Class 12 Engineering Chemistry Surface Chemistry Quiz 15 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Surface Chemistry
Quiz 15
Although nitrogen does not adsorb on surface at room temperature, it adsorbs on the same surface at
83K
. Which one of the following statements is correct?
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At 83 K, there is formation of monolayer
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At 83 K, nitrogen is adsorbed as atoms
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At 83 K, nitrogen molecules are held by chemical bonds
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At 83 K, there is formation of multimolecular layers
Read two statements:
(1) Milk is an example of oil in water (O/W) type emulsion
(2) Cold cream is an example of water in oil (W/O) type emulsion
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Only statement 1 is correct
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Only statement 2 is correct
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Both are correct
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None of these
Select correct statement:
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Water in oil emulsions are more viscous than the aqueous emulsions
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Electrical conductance of aqueous emulsions is less than that of oil emulsions
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De-emulsification can be done by soap or detergent
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An emulsion can be diluted with $$H_2O$$ then it is oil in water (O/W) type
$$10\%$$ sites of catalyst bed have adsorbed by $$H_{2}$$ on Heating $$H_{2} $$ gas is evolved from sites and collected at $$0.03$$ atm and $$300\,K$$ in a small vessel of $$2.46\,cm^{3}$$:
No. of sites available is $$5.4 \times 10^{16}\, per\,cm^{2} $$ and surface area is $$1000\,cm^{2}$$. Find out the no. of surfaces sites occupied per molecule of $$H_{2} $$. (Given $$N_{A} = 6 \times 10^{23})$$
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$$1$$
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$$2$$
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$$3$$
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None of these
Explanation
Adsorbed moles of $$H_{2} = \dfrac{0.03 \times 2.46 \times 10^{-3}}{0.0821 \times 300} $$
$$ = 3 \times 10^{-6} $$
$$ \therefore $$ No. of absorbed molecules of $$H_{2}$$
$$ = 3 \times 10^{-6} \times 6 \times 10^{23} $$
$$ \Rightarrow 18 \times10^{17} $$
Total no. of surface sites available
$$ = 5.4 \times 10^{16} \times 1000 $$
$$ \Rightarrow \, 5.4 \times 10^{19} \, cm^{2} $$
No. of surface sites that is occupied by adsorption of $$H_{2} $$
$$ = \dfrac{10}{100} \times 5.4 \times 10^{19}$$
$$ \Rightarrow $$ $$5.4 \times 10^{18} $$
No. of surface sites occupied by one molecule $$H_{2} $$
$$ = \dfrac{5.4 \times 10^{18}}{18 \times 10^{17}} = 3 $$
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