Question

What volume (in mL) of 18.0 Molarity H2SO4 is needed to contain 2.45 grams of H2SO4

Answer 1

1.39 mL

Step-by-step explanation:

To determine the volume of 18.0 Molarity
`H_(2) SO_4` needed to contain 2.45 grams of
`H_(2) SO_4`, we can use the following formula:

moles of solute = mass of solute / molar mass of solute

Then, we can use the molarity formula:

Molarity = moles of solute / volume of solution (in liters)

Rearranging this formula, we get:

volume of solution (in liters) = moles of solute / Molarity

First, we need to calculate the moles of
`H_(2) SO_4`:

moles of
`H_(2) SO_4` = mass of
`H_(2) SO_4` / molar mass of
`H_(2) SO_4`

The molar mass of
`H_(2) SO_4` is:

2(1.008 g/mol) + 32.06 g/mol + 4(16.00 g/mol) = 98.08 g/mol

Therefore, the moles of
`H_(2) SO_4` are:

moles of
`H_(2) SO_4` = 2.45 g / 98.08 g/mol = 0.02497 mol

Next, we can calculate the volume of 18.0 Molarity
`H_(2) SO_4` needed:

volume of solution (in liters) = moles of solute / Molarity

volume of solution (in liters) = 0.02497 mol / 18.0 mol/L = 0.001387 L

Finally, we can convert the volume to milliliters:

volume of solution (in mL) = 0.001387 L x 1000 mL/L = 1.39 mL

Therefore, approximately 1.39 mL of 18.0 Molarity
`H_(2) SO_4` is needed to contain 2.45 grams of
`H_(2) SO_4`.