What volume of 17 M sulfuric acid must be used to prepare 0.15 L of 1.78 M H2SO4?

M1V1=M2V2

17xV1=1.78X0.15
V1=0.01571 L

You are diluting the acid 17/1.78 times, or 9.55 times

That means one part acid, and 8.55 parts water.
What is one part? .15L/9.55=15.7ml
so, 8.55 parts water is 15.7*8.55=134.2 ml water.

so add 15.7 ml of 17M acid to the water, stirring, (always acid to water).

To find the volume of 17 M sulfuric acid needed to prepare 0.15 L of 1.78 M H2SO4, you can use the equation:

(moles of solute) = (molarity) x (volume)

Let's go step by step to find the solution:

1. Start by finding the moles of H2SO4 required in the final solution. Multiply the desired volume (in liters) by the desired molarity:

moles of H2SO4 = 0.15 L x 1.78 mol/L

2. Now, find the volume of 17 M sulfuric acid needed, using the moles of H2SO4 calculated in the previous step and the given concentration:

volume of 17 M sulfuric acid = (moles of H2SO4) / (concentration of sulfuric acid)

volume of 17 M sulfuric acid = (0.15 L x 1.78 mol/L) / (17 mol/L)

Simplify the equation:

volume of 17 M sulfuric acid = 0.02529 L (or 25.29 mL)

Therefore, you would need approximately 25.29 mL (or 0.02529 L) of 17 M sulfuric acid to prepare 0.15 L of 1.78 M H2SO4.