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?

ill get 30 extra points for these

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

Explanation:

Part A:

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

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

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

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.