what volume of 16 sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution

9.4 mL of acid

16 M*V=1.5 L * 0.10 M

V=(1.5 L * 0.10 M)/16M
=.00937L (9.375ml)

Assuming that's 16M, find volume V when:

(16M) * V = (0.10M) * (1.5L)

16 what? 16M?

c1v1 = c2v2
c = concn
v = volume
16M*v1 = 0.1*1.5M
Solve for v1.

Where is the work tho?

From dilution law

C1V1=C1V2
16M×V1=1.5M×0.10M
V1=0.00937L or 9.375mL

To determine the volume of 16 M sulfuric acid needed to prepare a 0.10 M H2SO4 solution, we can use the formula:

C1V1 = C2V2

where:
C1 = concentration of the initial solution
V1 = volume of the initial solution
C2 = concentration of the final solution
V2 = volume of the final solution

In this case, we know:
C1 = 16 M (the initial concentration of sulfuric acid)
V1 = unknown
C2 = 0.10 M (the desired final concentration of the H2SO4 solution)
V2 = 1.5 L (the desired final volume)

Plugging in the values into the formula:

(16 M) * V1 = (0.10 M) * (1.5 L)

To solve for V1, we rearrange the equation:

V1 = (0.10 M * 1.5 L) / 16 M

Calculating the value:

V1 = 0.015 L

Therefore, you would need to use 0.015 liters (or 15 milliliters) of 16 M sulfuric acid to prepare 1.5 L of a 0.10 M H2SO4 solution.