How many liters of water need to be added to 0.5 moles of CaCl2 to make a 0.20 M solution of CaCl2?
1 answer:
You might be interested in
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then,
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Above question is incomplete. Complete question is attached below
........................................................................................................................
Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case, reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is least reactive.
On other hand, reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is most reactive.
........................................................................................................................
Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case, reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is least reactive.
On other hand, reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is most reactive.