Question

If \[{\text{100 mL}}\] of \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] ​ and \[{\text{100 mL}}\]of \[{{\text{H}}_{\text{2}}}{\text{O }}\] are mixed, the mass percent of \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] in the resulting solution will be: \[({{\text{d}}_{{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}}} = 0.09{\text{g}}\,{\text{m}}{{\text{L}}^{{\text{ - 1}}}},{{\text{d}}_{{{\text{H}}_{{\text{2}}}}{\text{O}}}} = 1.0{\text{gm}}{{\text{L}}^{{\text{ - 1}}}})\]
A) $90.32\% $
B) $47.36\% $
C) $50.56\% $
D) $60\% $

Answer 1

Hint:
To answer this question, you should recall the method to find the percentage composition of an element in a compound. We shall find the mass of both the compounds and then find the percentage mass of sulphuric acid.

Formula used:
$\% {\text{mass of A = }}\dfrac{{{\text{mass of A}}}}{{{\text{mass of A + }}{\text{ mass of B}}}}$

Complete step by step solution:
The percentage composition refers to the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100.
According to this definition:
Mass of \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ }} = {\text{ }}\dfrac{{100 \times 0.9}}{{100 \times 0.9 + 100 \times 1}}{\text{ }} = {\text{ }}\dfrac{{90}}{{190}}{\text{ }} = {\text{ }}0.4736{\text{g}}.\]
$\therefore $Percentage Mass of \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} = 47.36\% \].

Hence, the correct answer to this question is option B.

Note:
Other concentration terms used are:
Concentration in Parts Per Million (ppm) The parts of a component per million parts (\[{10^6}\]) of the solution.
\[{\text{ppm(A) = }}\dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}}{ \times 1}{{\text{0}}^{\text{6}}}\]
Molality (m): Molality establishes a relationship between moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent. The molality formula is as given- \[{\text{Molality(m) = }}\dfrac{{{\text{Moles of solute}}}}{{{\text{Mass of solvent in kg}}}}\]
Normality: It is defined as the number of gram equivalents of solute present in one litre of the solution.
Mole Fraction: It gives a unitless value and is defined as the ratio of moles of one component to the total moles present in the solution. Mole fraction = $\dfrac{{{X_{\text{A}}}}}{{{X_{\text{A}}} + {X_B}}}$(from the above definition) where ${X_{\text{A}}}$is no. of moles of glucose and \[{X_{\text{B}}}\] is the no. of moles of solvent.