Question

Question #58bdc

Answer 1

You'd need 5.6 mL.

Explanation:

When dealing with such problems, so have to start from the target solution.

More specifically, you know that your target solution must have a volume of 1 L and a molarity of 0.1 M.

Since molarity is defined as moles of solute, in your case sulfuric acid, per liters of solution, you know that the target solution must contain 0.1 moles of sulfuric acid.

`C = n/V => n = C *V`

`n_(H_2SO_4) = "0.1 M" * "1 L" = "0.1 moles"`

Now, use sulfuric acid's molar mass to determine how many grams would contain this many moles

`0.1cancel("moles") * "98.08 g"/(1cancel("mole")) = "9.808 g"`

This is how much sulfuric acid you need your target solution to contain.

Now take a look at the stock solution. You know that it has a density of 1.80 g/mL. To make the calculations easier, assume that you have a 1-L sample of your stock solution.

This volume of solution would have a mass of

`1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.80 g"/(1cancel("mL")) = "1800 g"`

You also know that this stock solution is 98% w/w sulfuric acid, which means that you get 98 g of sulfuric acid for every 100 g of stock solution.

The mass of sulfuric acid present in your stock solution sample is

`1800cancel("g solution") * ("98 g"H_2SO_4)/(100cancel("g solution")) = "1764 g"`

So, if you get 1764 g of sulfuric acid for every 1 L of stock solution, you will need

`9.808cancel("g"H_2SO_4) * "1 L solution"/(1764cancel("g"H_2SO_4)) = "0.00556 L solution"`

I'll leave this rounded to two sig figs, despite the fact that you only gave one sig fig for the volume and molarity of the target solution. Expressed in mL, this is equivalent to

`0.00556cancel("L") * "1000 mL"/(1cancel("L")) = color(green)("5.6 mL")`

So, to prepare your target solution, you would need 5.6 mL of a 98% w/w stock solution of sulfuric acid.