2.00 L of 0.800 M #NaNO_3# must be prepared from a solution known to be 1.50 M in concentration. How many mL are required?

Question

Peyton Adams · Accepted Answer

1066.7mL of the 1.50M solution
or 1070 mL (3 sig figs)

The number of moles of solute, n, remains constant as a solution is diluted. #n_"initial" = n_"final"# or #C_"initial" xx V_"initial" = C_"final" xx V_"final"# #C_"initial" = 1.50 M# #V_"initial" = ? mL# #C_"final" = 0.800 M# #V_"final" = 2.00 L = 2000 mL# #V_"initial" = (C_"final" xx V_"final")/ C_"initial"# #V_"initial" = (0.800 xx 2000)/ 1.50# #V_"initial" = 1066.67 mL#

Avery Adams · Answer

undefined To prepare 2.00 L of 0.800 M NaNO3 from a solution known to be 1.50 M in concentration, you can use the dilution formula:

$$C_1V_1 = C_2V_2$$

Where:
- $C_1$ = initial concentration
- $V_1$ = initial volume
- $C_2$ = final concentration
- $V_2$ = final volume

Given:
- $C_1$ = 1.50 M
- $C_2$ = 0.800 M
- $V_2$ = 2.00 L

We need to find $V_1$, the initial volume of the solution to be added.

$$C_1V_1 = C_2V_2$$

$$1.50 \, \text{M} \times V_1 = 0.800 \, \text{M} \times 2.00 \, \text{L}$$

Solve for $V_1$:

$$V_1 = \frac{0.800 \, \text{M} \times 2.00 \, \text{L}}{1.50 \, \text{M}}$$

Calculate $V_1$ to find out how much of the 1.50 M solution is needed to prepare the desired concentration.

Then, convert the volume from liters to milliliters.