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Question

100 mL of sulphuric acid solution (sp. gr. =1.84) contains 98% by weight of pure acid. Calculate the volume of 0.46 M NaOH solution (in L) required to just neutralize the above acid solution.

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Solution

Density= mass/volume
mass=density×volume
given density of 98% H2SO4=1.84g/mL
1mL of 98% H2SO4 will weigh
1.84×1=1.84g
100mL will weigh ##1.84\times 100=184g$
concentration is given as 98% by mass, 184g of the solution will contain 0.98×184=180.32g of H2SO4
molar mass of H2SO4=98.1g
180.32g constitutes 1.84 moles
The reaction between NaOH and H2SO4 is as follows:
2NaOH+H2SO4Na2SO4+2H2O
, to neutralize 1 mole of H2SO4,
2 moles of NaOH are needed,
So, to neutralize 1.84 moles,
1.84×2=3.68 moles of NaOH
molar mass of NaOH=40g
So, 3.68 moles constitute 147g of NaOH
We know that 1L of 1M NaOH contains 40g of NaOH
So, 1L of 0.46M NaOH contains 18.4g of NaOH
to get 147g of NaOH, we need to take 147/18.4=7.826L of 0.1M NaOH
Volume of 0.46M NaOH solution required to neutralize the above acid solution is 7.826L

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