Molarity of NaNo by diluting 250.0 of a 1.60 solution to a final volume of 400cm3
original concentration is 1.6 mols in 1000cm^3
so mols in original 250 cc = (1/4)(1.6) = 0.4 mols
same mols in final
so
0.4 / 400 = M/1000
M = 400/400 = 1.00
Well, let's calculate the molarity of NaNo in the final solution by using the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, we need to find the moles of NaNo in the initial solution.
Moles = Molarity × Volume (in liters)
= 1.60 M × 0.250 L
= 0.4 moles
To calculate the molarity in the final solution, we need to convert the final volume from cm³ to liters.
Final volume = 400 cm³ = 400 / 1000 L
= 0.4 L
Now, we can use the formula to calculate the molarity in the final solution:
Molarity (M) = 0.4 moles / 0.4 L
= 1 M
So, the molarity of NaNo in the final solution is 1 M. But wait, what's NaNo? Sounds like a futuristic version of sodium nitrate! Maybe it's a new compound for clown chemistry experiments!
To determine the molarity of NaNO3 after dilution, we need to use the formula:
C1V1 = C2V2
Where:
C1 = initial molarity of the solution
V1 = initial volume of the solution
C2 = final molarity of the solution
V2 = final volume of the solution
Given:
C1 = 1.60 M
V1 = 250.0 mL (convert to L by dividing by 1000 mL/L)
V2 = 400 mL (convert to L by dividing by 1000 mL/L)
Converting V1 and V2 to liters:
V1 = 250.0 mL ÷ 1000 mL/L = 0.250 L
V2 = 400 mL ÷ 1000 mL/L = 0.400 L
Using the formula:
1.60 M (0.250 L) = C2 (0.400 L)
Solving for C2:
C2 = (1.60 M x 0.250 L) / 0.400 L
C2 = 1.00 M
Therefore, the molarity of NaNO3 after dilution is 1.00 M.
To find the molarity of NaNo in the final solution, we can use the formula:
M1V1 = M2V2
Where:
M1 = initial molarity of the solution (1.60 M)
V1 = initial volume of the solution (250.0 mL or 0.250 L)
M2 = final molarity of the solution (unknown)
V2 = final volume of the solution (400 mL or 0.400 L)
Now, we can rearrange the formula to solve for M2:
M2 = (M1 * V1) / V2
Plugging in the values we have:
M2 = (1.60 M * 0.250 L) / 0.400 L
M2 = 1.00 M
Therefore, the final molarity of NaNo in the solution is 1.00 M.