if a flask of 750 ml of 0.80m of nacl solution was left, and 275 ml evaporated, what is the new concentration
Since no NaCl was lost, we can use
C1V1=C2V2
C1=0.8 M
V1=750 mL
V2=(750-275)=475 mL
So
C2=C1V1/V2
=0.8*750/475
=1.26 mL
Note: I provide an extra significant figure when the first digit is "1". Your teacher may or may not recommend this.
I want to point out to you that you need to find the caps key on your keyboard and use it. You may (or may not) know that m stand for molality and M stands for molarity. Two separate animals. My assumption is that you meant M = molarrity and MathMate worked it that way, also. I also note that MathMate made a typo. That's 1.26 M for the concentration of the resulting solution.
Thank you for pointing out the oversight, and the delicate case difference!
To determine the new concentration after the evaporation, we can use the diluted solution formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration (what we want to find)
V2 = final volume
In this case, the initial concentration (C1) is 0.80 M and the initial volume (V1) is 750 ml. After 275 ml evaporated, the final volume (V2) will be 750 ml - 275 ml = 475 ml.
Now, let's substitute the given values into the formula and solve for C2:
C1V1 = C2V2
(0.80 M)(750 ml) = C2(475 ml)
To find C2, we rearrange the equation:
C2 = (C1V1) / V2
C2 = (0.80 M * 750 ml) / 475 ml
Now we can calculate the new concentration:
C2 = 1.28 M
So, the new concentration after the evaporation is approximately 1.28 M.