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Question: What is the purity of concentrated \[{H_2}S{O_4}\] solution (\[d = 1.8g/mol\] ) if \[5mL\] of these ...

What is the purity of concentrated H2SO4{H_2}S{O_4} solution (d=1.8g/mold = 1.8g/mol ) if 5mL5mL of these solution is neutralized by 84.5mL84.5mL of 2N$$$$NaOHsolution.
A. 93%93\%
B. 94.6%94.6\%
C. 92.12%92.12\%
D. 91.5%91.5\%

Explanation

Solution

The purity of a solution is the percentage purity of a substance which is evaluated by ratio of the mass of the pure substance and the total mass of the sample considered multiplied by 100.

Complete step by step answer:
The given reaction is a neutralization reaction between an acid and a base. The acid used is sulfuric acid and the base used is sodium hydroxide. The corresponding balanced reaction is as follows:
H2SO4+2NaOHNa2SO4+H2O{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + {H_2}O
The volume of NaOHNaOH solution used in the neutralization process is 84.5mL84.5mL. The Normality of the solution used is 2N2N. The moles of NaOHNaOH used is determined as
=2×84.51000=0.169moles= \dfrac{{2 \times 84.5}}{{1000}} = 0.169moles
Following the above equation it is clear that two equivalents of NaOHNaOH is used for consumption of 11 equivalent of H2SO4{H_2}S{O_4}. Thus for two moles of NaOHNaOH solution, one mole of H2SO4{H_2}S{O_4} is used for neutralization.
Hence for 0.169moles0.169moles of NaOHNaOH, moles of H2SO4{H_2}S{O_4} used are = 0.1692=0.0845moles.\dfrac{{0.169}}{2} = 0.0845moles.
A mole of substance is the ratio of the mass of the substance and the molar mass of the substance. Hence mole=mass of substancemolar mass of substancemole = \dfrac{{mass{\text{ }}of{\text{ }}subs\tan ce}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}substance}}
Molar mass of H2SO4{H_2}S{O_4} = 2 ×2{\text{ }} \times atomic mass of HH + atomic mass of SS + 4 ×4{\text{ }} \times atomic mass of OO
=2×1+32+4×16=98g/mol= 2 \times 1 + 32 + 4 \times 16 = 98g/mol
The mass of H2SO4{H_2}S{O_4} in 0.0845moles0.0845moles = moles of H2SO4{H_2}S{O_4} ×\times molar mass of H2SO4{H_2}S{O_4}.
=0.0845×98=8.281g= 0.0845 \times 98 = 8.281g
The total mass of H2SO4{H_2}S{O_4} which is used in the reaction is = density ×\times volume =1.8×5=9g = 1.8 \times 5 = 9g.
Hence the purity of the concentrated H2SO4{H_2}S{O_4} solution
=8.2819×100=92.01%= \dfrac{{8.281}}{9} \times 100 = 92.01\%
Thus option C is the correct answer.

Note: The purity or percentage of purity of a substance is determined using Quantitative chemistry calculations. The number of moles of the substance is the key to determine the purity of the substance.