The first ionisation potential of `Al` is smaller than that of `Mg` because:

Options

A. `20ml,10ml`
B. `100 ml, 60 ml`
C. `50ml 20 ml`
D. `50 ml, 35 ml`

Answer

1

Explanation

Final molarity `=(V_(1)xx1+V_(2)xx0.25)/((V_(1)+V_(2)))`
`0.75(V_(1)+V_(2))=V_(1)xx0.25`
`075V_(1)+0.75V_(2)=V_(1)+V_(2)xx0.25`
`0.5V_(2)=0.25V_(1)`
`(V_(1))/(V_(2))=2`