2.00L of 0.800M NaNO3 must be prepared froma solution known to be 1.50 in concentration. How many ml are required
To calculate the volume of the 1.50M NaNO3 solution required to prepare 2.00L of 0.800M NaNO3, we can use the dilution formula:
M1V1 = M2V2
Where:
M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume
Rearranging this formula to solve for V1, we get:
V1 = (M2 * V2) / M1
Plugging in the values:
M1 = 1.50M
M2 = 0.800M
V2 = 2.00L
V1 = (0.800M * 2.00L) / 1.50M
V1 = 1.067 L
To convert liters to milliliters, multiply by 1000:
V1 = 1.067 L * 1000 mL/L
V1 = 1067 mL
Therefore, 1067 mL of the 1.50M NaNO3 solution is required to prepare 2.00L of 0.800M NaNO3.
To calculate how many milliliters (mL) are required to prepare 2.00 L of a 0.800 M NaNO3 solution, you can use the equation:
C1V1 = C2V2
Where:
C1 = initial concentration of the known solution (1.50 M)
V1 = volume of the known solution (unknown)
C2 = desired concentration of the NaNO3 solution (0.800 M)
V2 = final volume of the NaNO3 solution (2.00 L)
Rearranging the equation to solve for V1, we have:
V1 = (C2V2) / C1
Substituting the known values into the equation:
V1 = (0.800 M x 2.00 L) / 1.50 M
V1 = (1.600 L) / 1.50
V1 ≈ 1.067 L (rounded to three decimal places)
To convert liters to milliliters, multiply the volume by 1000:
V1 ≈ 1.067 L x 1000 mL/L
V1 ≈ 1067 mL
Therefore, approximately 1067 mL (or 1067 milliliters) of the known solution are required to prepare 2.00 L of a 0.800 M NaNO3 solution.