The following volumes of 0.000300 M SCN are diluted to 15.00mL determine the concentration of SCN in each sample after dilution.
Sample 0.000300 M SCN(ml) SCN (M)
1 1.50
2 3.50
Can someone please explain how to do this?
You take 1.50 mL and dilute to 15.00; therefore, the new concn is
0.00030 x (1.50/15.00) = ??
If you prefer, you can use the dilution formula of c1v1 = c2v2 which is the same as mL1 x M1 = mL2 x M2
Should have been in a table but i guess it could not fit there.
Sample 1 = 1.50 (0.000300M SCN (mL)), SCN (M) = ?
sample 2 = 3.50 (0.000300M SCN (mL)), SCN (M) = ?
0.00003
Certainly! Let me break it down for you with a touch of humor.
To determine the concentration of SCN in each sample after dilution, we can use the formula: C1V1 = C2V2.
Now, for Sample 1, we have a concentration of 0.000300 M and a volume of 1.50 mL. After dilution, the volume becomes 15.00 mL, and we need to find the new concentration.
So, using the formula: (0.000300 M)(1.50 mL) = (C2)(15.00 mL).
Now, if we simplify that equation, we get: C2 = (0.000300 M)(1.50 mL) / (15.00 mL).
Calculating that out, the concentration of SCN in Sample 1 after dilution is... *drumroll*... tada... C2 = 0.000030 M.
Moving on to Sample 2, we have a concentration of 0.000300 M and a volume of 3.50 mL. Again, after dilution, the volume becomes 15.00 mL, and we need to figure out the new concentration.
Applying the formula: (0.000300 M)(3.50 mL) = (C2)(15.00 mL).
Simplifying that equation gives us: C2 = (0.000300 M)(3.50 mL) / (15.00 mL).
Doing the math, the concentration of SCN in Sample 2 after dilution is... *wait for it*... tada... C2 = 0.000070 M.
So, Sample 1 has a concentration of 0.000030 M SCN after dilution, and Sample 2 has a concentration of 0.000070 M SCN after dilution.
I hope that explanation tickled your funny bone and made the concept a little clearer!
To determine the concentration of SCN in each sample after dilution, you need to use the concept of dilution.
Dilution is the process of reducing the concentration of a solute in a solution by adding more solvent. It is governed by the equation:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, the initial volume is given as 0.000300 M SCN (ml). Let's calculate the final concentration for each sample:
Sample 1:
Given: Initial concentration (C1) = 0.000300 M
Initial volume (V1) = 1.50 ml
Final volume (V2) = 15.00 ml
Using the dilution equation:
C1V1 = C2V2
(0.000300 M)(1.50 ml) = C2(15.00 ml)
Rearranging the equation to solve for C2:
C2 = (0.000300 M)(1.50 ml) / (15.00 ml)
C2 = 0.000030 M
Therefore, the concentration of SCN in Sample 1 after dilution is 0.000030 M.
Sample 2:
Given: Initial concentration (C1) = 0.000300 M
Initial volume (V1) = 3.50 ml
Final volume (V2) = 15.00 ml
Using the dilution equation:
C1V1 = C2V2
(0.000300 M)(3.50 ml) = C2(15.00 ml)
Rearranging the equation to solve for C2:
C2 = (0.000300 M)(3.50 ml) / (15.00 ml)
C2 = 0.000070 M
Therefore, the concentration of SCN in Sample 2 after dilution is 0.000070 M.