Concentration

[Baria Alaswad 2E]

A solution contains 100.0mL of water and 0.70g NaCl. How much water needs to be added to dilute the concentration Cl- to one third of its original concentration?

[Leia1H]

For this, we can use the equation M1V1=M2V2
First, find how many grams of Cl you have, given the amount of NaCl. Then, convert the grams to moles. From here, you can find the initial concentration (M1) of the Cl. Now you have M1 and V1. M2 is stated to be 1/3M1. From here, plug the values into the equation above, and solve for V2, which should be the final volume needed for 1/3 concentration. For amount you need to add, just subtract the initial volume from the final. Hope this helps!

[AngeloCasubuan2B]

Hi Baria,

The first step would be to get the original molarity of the NaCl. To do this, convert the 100mL to L and the 0.70g NaCl to moles of NaCl.
100mL * 1L/1000mL = 0.100 L
0.70g NaCl * 1 mol NaCl / 58.443 g = 0.011978 mol NaCl
Molarity = mol/L -> 0.011978 mol NaCl / 0.100 L = 0.1198 M NaCl

The second step is to use the M1V1 = M2V2 equation, with M2 being M1 divided by 3, and solving for V2
V2 = (M1V1)/M2
((0.1198 M)(0.100L))/(0.1198/3 M) = 0.3L

Finally, subtract the original volume from the final volume to get how much water was added
0.300 L - 0.100 L = 0.200 L