If 4.0 L of a 4.8 M SrCl2 solution is diluted to 50 L , what is the molarity of the diluted solution?
4.8 M x (4.0/50) = ??
so would the answere be .43 M in two significant figures?
now i got a different answer .38 M
thank you .38 M is right
Its 0.38 !
Well, let's dive into the world of chemistry and dilutions! If we start with a 4.0 L solution of 4.8 M SrCl2 and dilute it to 50 L, we can calculate the molarity of the diluted solution.
Here's the math: M1V1 = M2V2
M1 = initial molarity = 4.8 M
V1 = initial volume = 4.0 L
M2 = final molarity (what we're trying to find)
V2 = final volume = 50 L
So, plugging the values in, we have:
(4.8 M)(4.0 L) = M2(50 L)
Solving for M2, we find:
M2 = (4.8 M * 4.0 L) / 50 L
M2 = 0.384 M
So, after all that chemistry number crunching, the molarity of the diluted solution is 0.384 M.
To find the molarity of a solution after dilution, you can use the dilution formula:
M1V1 = M2V2
Where:
M1 = initial molarity of the solution (before dilution)
V1 = initial volume of the solution (before dilution)
M2 = final molarity of the solution (after dilution)
V2 = final volume of the solution (after dilution)
In this case, the initial molarity of SrCl2 solution is 4.8 M and the initial volume is 4.0 L. The final volume after dilution is 50 L. We need to find the final molarity after dilution (M2).
Plugging in the values into the dilution formula:
(4.8 M)(4.0 L) = (M2)(50 L)
Solving for M2 by rearranging the equation:
M2 = (4.8 M)(4.0 L) / 50 L
M2 = (19.2 mol) / 50 L
M2 = 0.384 M
Therefore, the molarity of the diluted solution is 0.384 M.