"What volume of water must be added to 100ml of 15 M of H2SO4 to dilute to 5.0M?"

I've tried using lots of different methods but keep having the same answer and don't know if the answer book is wrong or if it's me. Thanks!

you are diluting by a factor of three (15/5). That means you need one part original, 2 parts water added.

One part=100ml of 15M
two parts water=200ml
result: 3parts (300ml) 5M H2SO4

First find the total volume by

V1C1 = V2C2
So, 0.1 x 15 = V2 x 5
We get V2 = 0.3L = 300mL

Now to find the amount of water required subtract the initial volume(V1) from the total volume V2

We get 300-100 = 200mL or 0.2L

Important: Don't ever add water in acid, always add acid in water. :D

To solve this dilution problem, you need to use the equation for dilution:

M1V1 = M2V2

Where:
M1 = Initial concentration
V1 = Initial volume
M2 = Final concentration
V2 = Final volume

In this case, you want to find the volume of water (V2) that needs to be added to 100 ml of 15 M H2SO4 (M1) in order to dilute it to 5.0 M (M2). Let's solve step by step:

Step 1: Convert the initial concentration and final concentration to moles per liter (mol/L):

Initial concentration (M1): 15 M
Final concentration (M2): 5.0 M

Step 2: Substitute the values into the dilution equation:

15 M * 100 ml = 5.0 M * (100 ml + V2)

Step 3: Solve for V2:

1500 ml = 500 ml + 5.0 M * V2
1000 ml = 5.0 M * V2
V2 = 1000 ml / 5.0 M

Step 4: Convert the volume from milliliters to liters:

V2 = 1000 ml / 5.0 M = 200 L

Therefore, you need to add 200 liters of water to 100 ml of 15 M H2SO4 to dilute it to 5.0 M.