0.200g of iodine is stirred in 100mL of water. After equilibrium is reached, we add 150mL of water to the system. How much iodine will be left undissolved?

0.200g of iodine is stirred in 100mL of water. After equilibrium is re

1.	0.200g of iodine is stirred in 100mL of water. After equilibrium is reached, we add 150mL of water to the system. How much iodine will be left undissolved?
Answer» `1.3g` `0.130g` `0.013g` `13g` Solution :After an equilibrium between `0.200g` of iodine and `100mL` of water is reached, the ADDITION of `150ML` of water will result in the further dissolution of `I_2`. VOLUME of solution after the addition of `150mL` of water `=100mL+150mL` `=250ML` According, `100mL` of solution contains `0.28g I_2`. `1mL` solution contains `(0.28g)/(1000ML)` Therefore, `250mL` of solution will contain `(0.28g)/(1000ml)xx250mL=0.070g` of `I_2` Undissolved iodine `=(0.200g)-(0.070g)` `=0.130g`