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Answer:
The new concentration of the solution is;
0.01 M
Explanation:
The question relates to concentration of solution and proportion of mixtures
The given parameters are;
The volume of water added to the 0.1 M solution = 900 mL
The initial volume of the solution to which water was added = 100 mL
Therefore, the total final volume of the solution = 100 mL + 900 mL = 1000 mL = 1 L
The concentration of the solution = 0.1 M
The number of moles present in the solution, 'n', is given as follows;
n = 100 mL/(1000 mL) × 0.1 M = 0.01 moles
Given that no more concentrated solution was added, we have;
The number of moles in the 100 mL solution = The number of moles in the 1,000 mL solution
Therefore, the number of moles in the 1,000 mL (1 L) solution = 0.01 moles
Therefore;
The new concentration of the 1,000 mL (1 L) solution = 0.01 moles/(1,000 mL) = 0.01 moles/(1 L) = 0.01 M (By definition of the molarity of a solution)
The new concentration of the 1,000 mL (1 L) solution = 0.01 M.
D. 0.01 M
Given:
The volume of water added to the 0.1 M solution = 900 mL
The initial volume of the solution to which water was added = 100 mL
So,
The total final volume of the solution = 100 mL + 900 mL = 1000 mL = 1 L
The concentration of the solution = 0.1 M
Firstly we need to find the number of moles which is represented by 'n':
[tex]n = \frac{100mL}{1000mL}* 0.1 M\\\\n = 0.01 \text{moles}[/tex]
The number of moles in the 100 mL solution = The number of moles in the 1,000 mL solution
Thus,
The number of moles in the 1,000 mL (1 L) solution = 0.01 moles
For calculation of new concentration:
Using Molarity formula:
The new concentration of the 1,000 mL (1 L) solution [tex]= \frac{0.01\text{ moles}}{1000mL} = \frac{0.01\text{ moles}}{1L} = 0.01 M[/tex]
The new concentration of the 1,000 mL (1 L) solution = 0.01 M.
Thus, the correct option is D.
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