If you take a 15.0-mL portion of the stock solution and dilute it to a total volume of 0.700L , what will be the concentration of the final solution?
Well, let's see... If you take a 15.0 mL portion of the stock solution and dilute it to a total volume of 0.700 L, you'll basically be stretching that solution, like a rubber band! So, if you think about it, the concentration of the final solution will be like the stamina of that rubber band - it'll be diluted!
But fear not, math is here to save the day. The formula we need is C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
So, in this case, we know that C1 is the concentration of the 15.0 mL stock solution portion (let's call it C1), V1 is 15.0 mL, C2 is the concentration we're looking for, and V2 is 0.700 L.
Now, we can plug in the values and solve for C2: C1V1 = C2V2. But be sure to convert that initial volume to liters before you calculate, otherwise, you'll end up with a really concentrated headache!
So go ahead, crunch those numbers, and let me know what hilarious concentration you end up with!
To find the concentration of the final solution, we can use the formula:
C1V1 = C2V2
Where:
C1 is the concentration of the stock solution
V1 is the volume of the stock solution used
C2 is the concentration of the final solution
V2 is the total volume of the final solution
Given information:
C1 = concentration of the stock solution
V1 = volume of the stock solution used = 15.0 mL = 15.0 mL * (1 L / 1000 mL) = 0.015 L
V2 = total volume of the final solution = 0.700 L
Now we can rearrange the formula to solve for C2:
C2 = (C1 * V1) / V2
Plugging in the values:
C2 = (C1 * 0.015 L) / 0.7 L
The concentration of the final solution will depend on the concentration of the stock solution (C1), which is not provided in the question. Please provide the concentration of the stock solution to calculate the concentration of the final solution.
To find the concentration of the final solution, we need to use the concept of molarity (M), which is defined as the number of moles of solute divided by the volume of solution in liters.
First, we need to determine the number of moles of solute in the 15.0 mL portion of the stock solution. This can be done by multiplying the volume by the concentration of the stock solution.
Let's assume the concentration of the stock solution is given in units of M (mol/L). If we denote the concentration of the stock solution as C_stock, then the number of moles of solute (n_stock) can be calculated as follows:
n_stock = C_stock * V_stock
In this case, V_stock is given as 15.0 mL. To convert it to liters, we divide by 1000: V_stock = 15.0 mL / 1000 mL/L = 0.015 L.
Now, we also need to determine the total volume of the final solution. From the problem statement, the total volume is given as 0.700 L.
Next, we use the moles of solute (n_stock) and the total volume of the final solution (V_final) to calculate the final concentration (C_final) using the formula:
C_final = n_stock / V_final
Substituting the values, we have:
C_final = (C_stock * V_stock) / V_final
Now we can determine the concentration of the final solution by plugging in the given values and performing the calculation.
Let M1 be the concentration of the stock solution.
Recall the formula relating molarity and volume:
M1 * V1 = M2 * V2
where
V1 = initial volume
V2 = final volume
M2 = final concentration
Substituting,
M1 * (15.0 mL) = (700 mL) * M2
We solve for M2:
M2 = M1 * 15/700
M2 = 0.0214 * M1
Hope this helps~ :3