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

Approximately 206 grams.

Step-by-step explanation:

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

`\text{Number of moles of solute} = \text{Concentration} * \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)* 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* \underbrace{1.008}_{\mathrm{H}} + 1* \underbrace{32.06}_{\mathrm{S}} + 4* \underbrace{15.999}_{\mathrm{O}} = \rm 98.072\;g\cdot mol^(-1)`.

`\text{Mass} = \text{Quantity in moles} * \text{Molar Mass}`.

`m = n \cdot M = \rm 2.10\; mol * 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.