78.7 mL sample of a 4.81 M sucrose solution is diluted to 120.0 mL. What is the molarity of the diluted soluti?
You can use the dilution formula of
c1v1 = c2v2 but I prefer reasoning. It must be 4.81 x dilution factor which is
4.81 x (78.7/120) = ?
To find the molarity of the diluted solution, we can use the equation:
M1 × V1 = M2 × V2
where
M1 = initial molarity of the solution
V1 = initial volume of the solution
M2 = final molarity of the solution (to be calculated)
V2 = final volume of the solution
In this case:
M1 = 4.81 M (initial molarity of the sucrose solution)
V1 = 78.7 mL (initial volume of the sucrose solution)
M2 = ?
V2 = 120.0 mL (final volume of the diluted solution)
Let's substitute these values into the equation and solve for M2:
(4.81 M) × (78.7 mL) = M2 × (120.0 mL)
Now, we can rearrange the equation to solve for M2:
M2 = (4.81 M × 78.7 mL) / 120.0 mL
M2 = 3.1196 M
Therefore, the molarity of the diluted solution is approximately 3.1196 M.