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Question: How many grams of \( {H_3}P{O_4} \) are in \( {\text{175 }}mL \) of a \( 3.5{\text{ }}M \) solution ...

How many grams of H3PO4{H_3}P{O_4} are in 175 mL{\text{175 }}mL of a 3.5 M3.5{\text{ }}M solution of H3PO4{H_3}P{O_4} ?

Explanation

Solution

Hint : The mole concept is very significant and useful in chemistry. It is actually the base of stoichiometry and it provides the best option to express the amounts of reactants as well as products that are consumed and formed during a chemical reaction.

Complete Step By Step Answer:
A solution's molarity tells you about the number of moles of solute that is present in 1 L of solution. You can say:
Molarity(M)=moles of solute1 litre of solution Molarity(M) = \dfrac{{moles{\text{ }}of{\text{ solute}}}}{{1{\text{ }}litre{\text{ of }}solution{\text{ }}}}
We know that in order to calculate a solution's molarity, you should know the number of moles of solute that is present in 1 L i.e. 1000 mL of solution.
We can use the molar mass of a compound to determine the number of moles present in the sample so number of moles of a compound can be calculated by applying the following formula:
Number of moles=Mass(g)Molar mass(gmol−1)Number{\text{ }}of{\text{ }}moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}mass(gmo{l^{ - 1}})}}
In the given question, we already know the molarity as well as volume of the solution which are as follows:
\begin{array}{*{20}{l}} {Molarity = 3.5} \\\ {Volume = 175mL = 0.175L} \end{array}
Substituting these given values in the formula of molarity, we will calculate the number of moles as depicted below:
3.5=moles of solute0.175  number of moles=0.6125 moles of H3PO4 \begin{gathered} 3.5 = \dfrac{{moles{\text{ }}of{\text{ }}solute}}{{0.175{\text{ }}}} \\\ number{\text{ }}of{\text{ }}moles = 0.6125{\text{ }}moles{\text{ }}of{\text{ }}{{\text{H}}_3}P{O_4} \\\ \end{gathered}
Molar mass of H3PO4{H_3}P{O_4} can be calculated by adding the atomic mass of each of the atom present in the molecule as shown below:
Molecular mass of H3PO4=(3×H)+(1×P)+(4×O)=(3×1)+(1×31)+(4×16)=98gmol−1Molecular{\text{ }}mass{\text{ }}of{\text{ }}{H_3}P{O_4} = (3 \times H) + (1 \times P) + (4 \times O) = (3 \times 1) + (1 \times 31) + (4 \times 16) = 98gmo{l^{ - 1}}
Now substituting the values in the number of moles formula, we get the mass of H3PO4{H_3}P{O_4} as shown below:
0.6125=Mass(g)98 Mass(g)=60.025g \begin{gathered} 0.6125 = \dfrac{{Mass(g)}}{{98}} \\\ Mass(g) = 60.025g \\\ \end{gathered}
Hence, 60.025 grams of phosphoric acid H3PO4{H_3}P{O_4} are in 175 mL{\text{175 }}mL of a 3.5 M3.5{\text{ }}M solution of H3PO4{H_3}P{O_4} .

Note :
Don’t get confused between molarity and molality. Molarity is the number of moles of solute per volume of solution (litres) while Molality is the number of moles of solute per weight of solvent (kilogram). These two have a relationship as below:
Molality = (density of the solution − molarity) × molecular weight of solute molarity.