Question

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

Answer 1

Answer:

The new concentration after addition of 900 mL of water to 100 mL of 0.1 M solution is 0.01 M.

Explanation:

We know that molarity M = number of moles of solute ÷ volume of solution in litres

Given Molarity = 0.1 M

Given initial volume = 100 mL = 0.1 L

⇒0.1 = n ÷ 0.1

⇒n = 0.1 × 0.1 = 0.01 moles

Now final volume = 900 + 100 = 1000 mL = 1 L

Final molarity = number of moles of solute ÷ final volume of solution in litres

                       = 0.01 ÷ 1 = 0.01 M

Therefore, the new concentration after addition of 900 mL of water to 100 mL of 0.1 M solution is 0.01 M.

Answer 2

Answer:

The new concentration of the solution is 0.01 M.

Explanation:

M1 × V1 = M2 × V2

Using the equation, we can find the concentration or volume of the concentrated or dilute solution.

M1 = 0.1 M

V1 = 100 mL

V2 =1000 mL( That is, 900 mL water added to 100 mL of the solution. Therefore the total volume becomes 1000 mL)

M2 =?

M1 × V1 = M2 × V2

∴ M2 = $\frac{M1 * V1}{V2}$ = $\frac{0.1*100}{1000}$ = 0,01 M

That is the new concentration of the solution is 0.01 M.