What volume of 6.48 M HNO3 is required to

prepare 301 mL of a 1.2 M HNO3 solution?
Answer in units of mL.

mL x M = mL x M

To find the volume of a solution needed, we can use the formula:

(M1)(V1) = (M2)(V2)

where M1 and V1 represent the concentration and volume of the initial solution, and M2 and V2 represent the concentration and volume of the final solution.

Let's plug in the given values:

M1 = 6.48 M (concentration of the initial solution)
V1 = ? (unknown volume of the initial solution)
M2 = 1.2 M (concentration of the final solution)
V2 = 301 mL (volume of the final solution)

Now, rearrange the formula to solve for V1:

V1 = (M2 * V2) / M1

Substituting the values:

V1 = (1.2 M * 301 mL) / 6.48 M

V1 = 55.56 mL

Therefore, the volume of 6.48 M HNO3 required to prepare 301 mL of a 1.2 M HNO3 solution is 55.56 mL.