The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
Answer:
Option (B) is correct
Explanation:
Oxidation:
Reduction:
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Overall:
Nernst equation for this cell reaction at :
Where, n is number of electron exchanged and species inside third bracket represents concentrations in molarity.
Here, n = 2, and
So, plug in all the given values into above equation:
So,
As the value "0.059" varies from literature to literature and is most closest to therefore option (B) is correct.
Answer:
The answer is boiling point