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The density of solid argon is `1.65 "g mL"^(-1)` at `-233^@C` . If the argon atom is assumed to be sphere of radius `1.54xx10^(-8)` cm, what percentage of solid argon is apparently empty space ? (At. Mass of Ar =40) |
Answer» Density of `1.65 "g mL"^(-1)` means that 1.65 g of solid argon has volume =1 mL Volume of one atom of Ar =`4/3pir^3` No. of atom in 1.65 g or 1 mL of solid Ar=`(1.65/40"mol")(6.023xx10^23 "mol"^(-1))=(1.65xx6.023xx10^23)/40` `therefore` Total volume of all atoms Ar in 1 mL of Ar =`4/3pir^3 xx (1.65xx6.023xx10^23)/40` `=4/3xx22/7xx(1.54xx10^(-8))^3 xx (1.65xx6.023xx10^23)/40` `=0.380 cm^3` or 0.380 mL Volume of solid Ar = 1 mL `therefore` Empty space =(1-0.380) mL =0.620 mL `therefore` % empty space =`0.620/1xx100` =62% |
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