Question

When ${{12}}{{.044 \times 1}}{{{0}}^{{{23}}}}$molecule of ${{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}$ is dissolve in ${{195c}}{{{m}}^3}$ solution. What is mass percent ${{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}$ in the solution. If the density of the solution is ${{1}}{{.25gc}}{{{m}}^{{{ - 1}}}}$.

Answer 1

Percent composition describes percentage of each component in the given solution. The composition of the solution can be described in mass percentage. It is the mass of solute present in each mass of solution. And it describes the percentage of an element in the total compound.

Complete step by step answer:
The atomic mass of an element is the mass of one atom of the element in terms of atomic mass units (amu). It gives an average value to the mass of one atom of the element.
The molar mass of a substance is the total mass of one mole of the substance. It is expressed in terms of ‘grams per mole $\left( {{{gmo}}{{{l}}^{ - 1}}} \right)$. The gram atomic mass (GAM) of an element is the mass of one mole of that element and the gram molecular mass (GMM) of a compound refers to the mass of one mole of the compound.
For instance, the gram atomic mass of hydrogen is approximately ${{1}}{{.007g}}$ and the gram molecular mass of water is ${{18}}{{.015g}}$.
The number of moles of an element or compound, ${{n}}$ can be calculated by dividing the total mass of the sample, ${{{m}}_{{s}}}$ by the molar mass of the element or compound, ${{M}}$, as described by the following formula.
Number of Moles, ${{n}} = \dfrac{{{{{m}}_{{s}}}}}{{{M}}}$
The total number of atoms or molecules in a sample is obtained by multiplying the number of moles with the Avogadro number.
Number of Atoms or Molecules ${{ = }}\left( {{n}} \right){{ \times }}\left( {{{6}}{{.022 \times 1}}{{{0}}^{{{23}}}}} \right)$
Calculation: The density of solution is given by ${{1}}{{.25gc}}{{{m}}^{ - 1}}$. It is the ratio of mass to density
Density $\rho {{ = \;\;}}\dfrac{{{m}}}{{{V}}}$
${{m = }}\rho \times {{V}}$ $\Leftrightarrow {{1}}{{.25 \times 195 = 243}}{{.75g}}$
Number of molecules of sulphuric acid, ${{{n}}_{{{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}}} = 1{{2}}{{.044 \times 1}}{{{0}}^{{{23}}}}$
${{{n}}_{{{{H}}_2}{{S}}{{{O}}_4}}}{{ = }}\,\,\dfrac{{{{{m}}_{{{{H}}_2}{{S}}{{{O}}_4}}}}}{{{M}}}{{ \times }}{{{N}}_{{A}}}$
${{{m}}_{{{{H}}_2}{{S}}{{{O}}_4}}}{{ = }}\dfrac{{{{12}}{{.044}} \times {{10}^{23}}{{ \times 108}}}}{{{{6}}{{.022 }} \times {{ }}{{10}^{23}}}} = 216{{g}}$
${{{m}}_{^{{{{H}}_2}{{S}}{{{O}}_4}}}}{{ = }}\dfrac{{{{216 \times 100}}}}{{{{243}}{{.75}}}}{{ = 88}}{{.61\% = 89\% }}$
Hence, the answer to the above question is ${{89\% }}$

Note: Molarity and molality is also a measure of concentration of a solution. Molarity is the number of moles of solute per litre of solution. It is denoted by the symbol M. Molality can be defined as the number of moles of solute in one kilogram of solvent.