A) 0.0355 g
B) 0.071 g
C) \[\text{6}.0\text{23 }\times \text{1}{{0}^{\text{21}}}\]molecules
D) 0.02 mol
Correct Answer: A
Solution :
Ideal gas equation is \[pV=nRT\] When V and T are same, \[p\propto n\] Thus, when number of moles, i.e., n is least, it will exert least pressure. \[\text{n =}\frac{\text{wt}}{\text{mol}\text{.wt}}\text{=}\frac{\text{0}\text{.0355}}{\text{35}\text{.5}}\text{=1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{mol}\] \[\text{n =}\frac{\text{0}\text{.071}}{\text{35}\text{.5}}\text{=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{mol}\] \[\text{n =}\frac{\text{number or molecules}}{{{\text{N}}_{\text{A}}}}\] \[\text{=}\frac{\text{6}\text{.023 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}}{\text{6}\text{.023 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}\text{=0}\text{.01 mol}\] \[\text{n = 0}\text{.02 mol}\] Thus, 0.0355 g chlorine will exert the least pressure.
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