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`0.1 M` aqueous solution of `MgCl_(2)` at `300 K` is `4.92 atm`. What will be the percentage ionination of the salt? a.`49%` , b. `59%` ,c.`79%` d. `69%` |
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Answer» a.`pi=iCRT` `4.92=ixx0.1xx0.0821xx300` `i=1.99` `alpha=(i-1)/(n-1)=(1.99-1)/(3-1)=(0.99)/(2)=0.49` Perecentage of ionization=`49%` |
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