Question

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

Answer 1

Answer:

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

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):

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

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.

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

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)

$\displaystyle V = \frac{n}{c} = \dfrac{5.0983 \times 10^{-4}}{0.0510} \approx \rm 1.00\times 10^{-2}\; L$.

That's the same as $\rm 1.00\times 10^{-2} \times 10^{3} \; mL = 10.0\; mL$.