Use the drop-down menus to identify the class of compounds to which each structure belongs.
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seaweed having 50 g salt in 1 L water. The bucket contains 150 g of salt in 2 L of water
amount of water present in bucket is twice to amount of water in weed
At equilibrium, volume of water in weed is x and volume in bucket is y but concentration remain same as follows:
At equilibrium, weed loose z L from 1 L water to bucket containing 2 L as follows:
Thus, Weed will loose 0.25 L of water
Explanation:
The given reaction will be as follows.
............. (1)
=
Reaction for the complex formation is as follows.
........... (2)
=
When we add both equations (1) and (2) then the resultant equation is as follows.
............. (3)
Therefore, equilibrium constant will be as follows.
K =
=
=
Since, we need 0.010 mol of AgCl to be soluble in 1 liter of solution after after addition of for complexation. This means we have to set
=
=
= 0.010 M
For the net reaction,
Initial : 0.010 x 0 0
Change : -0.010 -0.020 +0.010 +0.010
Equilibrium : 0 x - 0.020 0.010 0.010
Hence, the equilibrium constant expression for this is as follows.
K =
=
x = 0.0945 mol
or, x = 0.095 mol (approx)
Thus, we can conclude that the number of moles of needed to be added is 0.095 mol.