Question

A chemist requires 5.00 liters of 0.420 M H2SO4 solution. How many grams of H2SO4 should the chemist dissolve in water? 129 grams 161 grams 206 grams 226 grams

Answer 1

Answer:

Approximately 206 grams.

Explanation:

How many moles of sulfuric acid $\mathrm{H_2SO_4}$ are there in this solution?

$\text{Number of moles of solute} = \text{Concentration} \times \text{Volume}$.

The unit for concentration "$\mathrm{M}$" is equivalent to mole per liter. In other words, $\rm 1\;M = 1\; mol\cdot L^{-1}$. For this solution, the concentration of $\mathrm{H_2SO_4}$ is $\rm 0.420\;M = 0.420\; mol\cdot L^{-1}$.

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V\\&= \rm 0.420\;mol\cdot L^{-1}\times 5.00\; L \\&= \rm 2.10\; mol\end{aligned}$.

What's the mass of that $\rm 2.10\; mol$ of $\mathrm{H_2SO_4}$?

Start by finding the molar mass $M$ of $\mathrm{H_2SO_4}$.

Relative atomic mass data from a modern periodic table:

• H: 1.008;
• S: 32.06;
• O: 15.999.

$\displaystyle M(\mathrm{H_2SO_4}) = 2\times \underbrace{1.008}_{\mathrm{H}} + 1\times \underbrace{32.06}_{\mathrm{S}} + 4\times \underbrace{15.999}_{\mathrm{O}} = \rm 98.072\;g\cdot mol^{-1}$.

$\text{Mass} = \text{Quantity in moles} \times \text{Molar Mass}$.

$m = n \cdot M = \rm 2.10\; mol \times 98.072\;g\cdot mol^{-1} \approx 206\; g$.

In other words, the chemist shall need approximately 206 grams of $\mathrm{H_2SO_4}$ to make this solution. As a side note, keep in mind that the 206 grams of $\mathrm{H_2SO_4}$ also take up considerable amount of space. Therefore it will take less than 5.00 L of water to make the 5.00 L solution.