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A mixture of `NO_(2) and N_(2)O_(4)` has a vapour density of `38*3` at 300 K . What is the number of moles of `NO_(2)` in 100 g of the mixture ?A. `0*043`B. `4*4`C. `3*4`D. ` 0*437` |
Answer» Correct Answer - D Suppose `NO_(2) = x g . " Then " N_(2) O_(4) = ( 100 - x) g ` `" Moles of " NO_(2)= x/46,"Moles of " N_(2)O_(4)=(100-x)/92` `"Mole fraction of "NO_(2)(x//46)/(x//46 + ( 100 - x ) //92)` `= x/46 xx92/(100+x)=(2x)/(100+x)` Mole fraction of `N_(2)O_(4) = 1 - (2x)/(100+x) = (100 - x)/(100+x) ` Molar mass of mixture `= (2x)/(100 + x) xx 46 + ( 100-x)/(100+x) xx92 = 9200/(100+x) ` ` :. 9200/(100+x) = 2 xx 38*3 = 76*6` or` 76*6 x = 9200 = 1540 or x = 20*10 g ` ` :." Moles of " NO_(2) = (20*10)/46 = 0. 437` |
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