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0.1 M aqueous solution of MgCl(2) at 300...

`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%`

A

0.49

B

0.29

C

0.39

D

0.69

Text Solution

Verified by Experts

The correct Answer is:
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`
Percentage ionization = 49%]
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