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0.200g of iodine is stirred in 100mL of ...

`0.200g` of iodine is stirred in `100mL` of water at `298K` till equilibrium is reached:
`[I_2(aq.)]_("Equilibrium")=0.0011 mol L^(-1)` at `298K`
What will be the mass of iodine found in solution?

A

`0.028g`

B

`0.28g`

C

`0.0028g`

D

`2.8g`

Text Solution

Verified by Experts

The correct Answer is:
A
Molar mass of iodine `I_2=254 g mol^(-1)`
Thus, the mass of iodine dissolved in `1L (1000mL)` of water at equilibrium is
`=`(Molarity) x (Molar mass)
`=(0.0011 mol L^(-1))(254 g mol^(-1))`
`=0.2794 g L^(-1)`
`=0.28g L^(-1)`
Thus, the mass of dissolved iodine in `100mL=0.028g`.
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