900 mL of water is added to 100 mL of a 0.1 M solution. What is the new concentration? 0.1M
0.001M
0.0001M
0.01M

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':

 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

The new concentration of the 1,000 mL (1 L) solution = 0.01 M.

Thus, the correct option is D.

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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.

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