How many mL of 0.25 M H2SO4 can be made from 50.0mL of 0.50 M H2SO4?
Also in this case, what does the capital M stand fro?
M = molarity = moles/L soln.
mL x 0.25M = 50.0 mL x 0.50 M
Well, here's a chemistry joke for you: Why did the scarecrow become a chemist? Because he loves chemistry...especially that straw-nium!
Now let's get to your question. The capital M stands for "molarity," which is a unit used to measure the concentration of a solution. In this case, 0.50 M H2SO4 means that there are 0.50 moles of sulfuric acid in every liter (1000 mL) of solution.
To find out how many mL of 0.25 M H2SO4 can be made from 50.0 mL of 0.50 M H2SO4, we need to use the concept of moles and molarity.
First, let's find the number of moles in the given 50.0 mL of 0.50 M H2SO4 solution.
Moles = Molarity * Volume (in liters)
Moles = 0.50 moles/L * 0.050 L = 0.025 moles
Now, we can use stoichiometry to find the volume of 0.25 M H2SO4 solution that can be made.
Moles = Molarity * Volume (in liters)
0.025 moles = 0.25 moles/L * Volume (in liters)
Solving for Volume (in liters):
Volume (in liters) = 0.025 moles / (0.25 moles/L)
Volume (in liters) = 0.1 L
Finally, we can convert the volume from liters to milliliters:
Volume (in mL) = 0.1 L * 1000 mL/L
Volume (in mL) = 100 mL
Therefore, 100 mL of 0.25 M H2SO4 can be made from 50.0 mL of 0.50 M H2SO4.
To find out how many mL of 0.25 M H2SO4 can be made from 50.0 mL of 0.50 M H2SO4, we can use the formula for dilution:
C1V1 = C2V2
where:
C1 = concentration of initial solution
V1 = volume of initial solution
C2 = concentration of final solution
V2 = volume of final solution
Let's plug in the values given:
C1 = 0.50 M
V1 = 50.0 mL
C2 = 0.25 M
V2 = ? (unknown)
Rearranging the formula, we get:
V2 = (C1V1) / C2
Substituting in the values, we have:
V2 = (0.50 M)(50.0 mL) / 0.25 M
Simplifying the equation:
V2 = 1000 mL / 0.25
V2 = 4000 mL
Therefore, 4000 mL of 0.25 M H2SO4 can be made from 50.0 mL of 0.50 M H2SO4.
In this case, the capital "M" stands for molarity, which is a unit of concentration. It represents the number of moles of a solute dissolved in one liter of solution.
To find the number of milliliters (mL) of a 0.25 M H2SO4 solution that can be made from a 50.0 mL of 0.50 M H2SO4 solution, we can use the concept of dilution.
Dilution is a process of reducing the concentration of a solute in a solution by adding more solvent. The relationship between the initial concentration, volume, and final concentration of a solution can be given by the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution 1
V1 = initial volume of the solution 1
C2 = final concentration of the solution 2
V2 = final volume of the solution 2
In this case, we have:
C1 = 0.50 M (initial concentration, 0.50 moles per liter)
V1 = 50.0 mL (initial volume, 50.0 milliliters)
C2 = 0.25 M (final concentration, 0.25 moles per liter)
We need to find V2, the final volume of the 0.25 M H2SO4 solution that can be made.
Using the formula mentioned earlier, we can rearrange it to solve for V2:
V2 = (C1 * V1) / C2
Plugging in the values:
V2 = (0.50 M * 50.0 mL) / 0.25 M
Now, let's calculate V2:
V2 = (0.50 * 50.0) / 0.25 = 100 mL
Therefore, 100 mL of a 0.25 M H2SO4 solution can be made from 50.0 mL of a 0.50 M H2SO4 solution.
As for the capital "M" in the unit of concentration, it stands for molarity. Molarity is a measure of the concentration of a solution and represents the number of moles of a solute in one liter of the solution. In this case, a 0.50 M H2SO4 solution means that there are 0.50 moles of H2SO4 dissolved in one liter of the solution. Similarly, a 0.25 M H2SO4 solution means that there are 0.25 moles of H2SO4 dissolved in one liter of the solution.