Respuesta :
w=0.254 (25.4%)
M(H₃PO₄)=98.0 g/mol
m - the mass of the solution
m(H₃PO₄)=mw
n(H₃PO₄)=m(H₃PO₄)/M(H₃PO₄)
m(H₂O)=m(1-w)
the molality is:
Cm=n(H₃PO₄)/m(H₂O)
Cm=mw/[M(H₃PO₄)m(1-w)]=w/[M(H₃PO₄)(1-w)]
Cm=0.254/[98*(1-0.254)]=0.003474 mol/g=3.474*10⁻³ mol/g=3.474 mol/kg
M(H₃PO₄)=98.0 g/mol
m - the mass of the solution
m(H₃PO₄)=mw
n(H₃PO₄)=m(H₃PO₄)/M(H₃PO₄)
m(H₂O)=m(1-w)
the molality is:
Cm=n(H₃PO₄)/m(H₂O)
Cm=mw/[M(H₃PO₄)m(1-w)]=w/[M(H₃PO₄)(1-w)]
Cm=0.254/[98*(1-0.254)]=0.003474 mol/g=3.474*10⁻³ mol/g=3.474 mol/kg
The molality of a 25.4 % (by mass) aqueous solution of phosphoric acid [tex]\left({{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}}\right)[/tex] is [tex]\boxed{{\text{3}}{\text{.47 m}}}[/tex] .
Further Explanation:
The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:
1. Molarity (M)
2. Molality (m)
3. Mole fraction (X)
4. Parts per million (ppm)
5. Mass percent ((w/w) %)
6. Volume percent ((v/v) %)
Molality is equal to the amount of solute that is dissolved in one kilogram of the solvent. It is denoted by m and its unit is moles per kilograms (mol/kg). The component present in smaller quantity is solute while that in larger quantity is known as a solvent. The solute gets itself dissolved in the solvent.
The formula to calculate the molality of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] solution is as follows:
[tex]{\text{Molality of }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}\;{\text{solution}}=\frac{{{\text{Amount}}\;\left({{\text{mol}}} \right)\;{\text{of}}\;{{\text{H}}_3}{\text{P}}{{\text{O}}_4}}}{{{\text{Mass}}\;\left({{\text{kg}}}\right)\;{\text{of}}\;{{\text{H}}_2}{\text{O}}}}[/tex] …… (1)
Consider 100 g to be the mass of the sample. Phosphoric acid [tex]\left({{{\text{H}}_3}{\text{P}}{{\text{O}}_4}}\right)[/tex] is 25.4 % by mass. So the mass of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] is 25.4 g.
The formula to calculate the mass of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] solution is as follows:
[tex]{\text{Mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ solution}}={\text{Mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}+{\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] …… (2)
Rearrange equation (2) to calculate the mass of [tex]{{\text{H}}_2}{\text{O}}[/tex] .
[tex]{\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O}}={\text{Mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ solution}}-{\text{Mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] …… (3)
The mass of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] solution is 100 g.
The mass of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] is 25.4 g.
Substitute these values in equation (3).
[tex]\begin{aligned}{\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O}}&={\text{100 g}}-{\text{25}}{\text{.4 g}}\\&={\text{74}}{\text{.6 g}}\\\end{aligned}[/tex]
Mass of [tex]{{\text{H}}_2}{\text{O}}[/tex] is to be converted into kg. The conversion factor for this is,
[tex]{\text{1 g}}={10^{ - 3}}\;{\text{kg}}[/tex]
So the mass of [tex]{{\text{H}}_2}{\text{O}}[/tex] is calculated as follows:
[tex]\begin{aligned}{\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O}}&=\left({{\text{74}}{\text{.6 g}}}\right)\left({\frac{{{{10}^{ - 3}}\;{\text{kg}}}}{{{\text{1 g}}}}}\right)\\&=0.074{\text{6 kg}}\\\end{aligned}[/tex]
The formula to calculate the moles of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] is as follows:
[tex]{\text{Moles of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}=\frac{{{\text{Given mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}}}{{{\text{Molar mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}}}[/tex] …… (4)
Substitute 25.4 g for the given mass and 97.99 g/mol for the molar mass of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] in equation (4).
[tex]\begin{aligned}{\text{Moles of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}&=\left( {{\text{25}}{\text{.4 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{97}}{\text{.99 g}}}}}\right)\\&=0.259{\text{2 mol}}\\\end{aligned}[/tex]
Substitute 0.2592 mol for the amount of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] and 0.0746 kg for the mass of [tex]{{\text{H}}_2}{\text{O}}[/tex] in equation (1).
[tex]\begin{aligned}{\text{Molality of }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}\;{\text{solution}}&=\frac{{{\text{0}}{\text{.2592 mol}}}}{{{\text{0}}{\text{.0746 kg}}}}\\&=3.4746{\text{6 m}}\\&\approx 3.4{\text{7 m}}\\\end{aligned}[/tex]
Therefore the molality of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] solution is 3.47 m.
Learn more:
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Solutions
Keywords: concentration terms, molality, m, mol/kg, H3PO4, moles of H3PO4, mass of H2O, H2O, 25.4 %, 25.4 g, 100 g, 74.6 g, 0.2592 mol, 3.47 m.
