We’re being asked to calculate the mass percent of H_{3}PO_{4} (MM = 98) in a 2.00 m aqueous H_{3}PO_{4} solution.

*Aqueous solution** means that **water is the solvent**, hence **CdSO _{4} is the solute*

**Mass** or **weight percent (% by mass) **is the percentage of a given element or compound within a solution.

The equation used to calculate for mass percent is shown below:

$\overline{){\mathbf{m}}{\mathbf{a}}{\mathbf{s}}{\mathbf{s}}{\mathbf{}}{\mathbf{p}}{\mathbf{e}}{\mathbf{r}}{\mathbf{c}}{\mathbf{e}}{\mathbf{n}}{\mathbf{t}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{n}\mathbf{e}\mathbf{n}\mathbf{t}}{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{\times}}{\mathbf{100}}}$

Which of the following represents the mass percent of H_{3}PO_{4} (MM = 98) in a 2.00 m aqueous H_{3}PO_{4} solution?

A. 19.6%

B. 16.4%

C. 9.8%

D. 8.2%

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