Question

What volume of 0.0510 M. H2SO4 solution contains 50.0 mg solute

Answer 1

Approximately 0.0100 L. That's equivalent to 10.0 mL.

Step-by-step explanation:

Look up the relative atomic mass data from a modern periodic table:

• H: 1.008;
• S: 32.06;
• O: 15.999.

Calculate the molar mass of
`\rm H_2SO_4` (sulfuric acid):

`M\left(\mathrm{H_2SO_4}\right) = 2 * 1.008 + 32.06 + 4 * 15.999 = \rm 98.072\; g \cdot mol^(-1)`.

The standard unit of mass is gram. To make calculations easier, convert the mass of
`\rm H_2SO_4` to that unit:

`m(\mathrm{H_2SO_4}) = \rm 50.0\;mg = 0.0500\; g`.

Find the number of moles of
`\rm H_2SO_4` in this solution. Both the mass and the molar mass are in their respective standard unit. As a result, the value from this calculation will also be in the appropriate standard unit.

`\displaystyle n(\mathrm{H_2SO_4}) = (m)/(M) = (0.0500)/(98.072) \approx \rm 5.0983 * 10^(-4)\; mol`.

Calculate the volume of this solution. Note that for concentration,
`\rm 1\;M` is the same as
`\rm 1\; mol \cdot L^(-1)` (moles per liter, not per milliliters)

`\displaystyle V = (n)/(c) = (5.0983 * 10^(-4))/(0.0510) \approx \rm 1.00* 10^(-2)\; L`.

That's the same as
`\rm 1.00* 10^(-2) * 10^(3) \; mL = 10.0\; mL`.