Answer: [tex]\dfrac{11}{9}I[/tex]
Step-by-step explanation:
Given
There is [tex]l[/tex] liter of pure alcohol
Suppose [tex]x[/tex] liters of water is added
After addition of water, alcohol becomes 45% in concentration
[tex]\Rightarrow \dfrac{l}{x+l}=45\%\\\\\Rightarrow \dfrac{I}{0.45}=x+I\\\\\Rightarrow \dfrac{20}{9}I-I=x\\\\\Rightarrow x=\dfrac{11}{9}I[/tex]
Thus, [tex]\dfrac{11}{9}I[/tex] of water is added to the pure alcohol.