Question

2.00L of 0.800M NaNO3 must be prepared froma solution known to be 1.50 in concentration. How many ml are required

Answer 1

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.