Question

A. A 3.24 M solution of base is Diluted by adding water increasing its volume from 1.25 L to 3.98 L. Calculate the new Molarity of the solution

B. 270 mL of a 7.0 M NaHCO3 solution was used in the preparation of 1.9 L of diluted NaHCO3. what is the molarity of the diluted solution

C. what volume of 5 M NaOH is needed to create a 100 mL solution of 1 M NaOH?

PLEASE HELP ILL GET 30 EXTRA POINTS FOR THESE

Answer 1

A. The new molarity of the diluted solution is 1.0176 M

B. The new molarity of the diluted solution is 0.9947 M

C. 20 mL of volume is required to prepare 100ml of 1 M NaOH

Step-by-step explanation:

Part A:

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This problem will be solved by using the following formula

M₁V₁ = M₂V₂ (1)

Given Data:

Initial Molarity of Solution M₁ = 3.24 M

Initial volume V₁ = 1.25 L

Initial volume V₂ = 3.98 L

Final Molarity M₂= ?

Put the values in formula 1

3.24 M x 1.25 L = M₂ x 3.98 L

By Rearranging

M₂ = 3.24 M x 1.25 L / 3.98 L

M₂ = 4.05 / 3.98

M₂ = 1.0176 M

The new molarity of the diluted solution is 1.0176 M

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Part B:

This problem will be solved by using the following formula

M₁V₁ = M₂V₂ (1.1)

Given Data:

Initial Molarity of Solution M₁ = 7.0 M

Initial volume V₁ = 270 mL

As the initial volume is in ml and final volume in Litre (L)

So we have to convert mL to L

1000ml= 1L

270 ml = ? L

1 x 270/ 1000 = 0.27 L

So

Initial volume V₁ = 0.27 L

Initial volume V₂ = 1.9 L

Final Molarity M₂= ?

Put the values in formula 1.1

M₁V₁ = M₂V₂

7.0 M x 0.27 L = M₂ x 1.9 L

By Rearranging

M₂ = 7.0 M x 0.27 L / 1.9 L

M₂ = 1.89 / 1.9

M₂ = 0.9947 M

The new molarity of the diluted solution is 0.9947 M

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Part 3

This problem will be solved by using the following formula

M₁V₁ = M₂V₂ (1.2)

Given Data:

Initial Molarity of Solution M₁ = 5 M

Initial volume V₁ = ? mL

Initial volume V₂ = 100 mL

Final Molarity M₂= 1 M

Put the values in formula 1.2

M₁V₁ = M₂V₂

5 M x V₁ = 1 M x 100 mL

By Rearranging

V₁ = 1 M x 100 mL/ 5 M

V₁ = 100 / 5

V₁ = 20

20 mL of volume is required to prepare 100ml of 1 M NaOH