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Concentration Terms - An Introduction (w/w etc)

A 1.5 g sam...

Question

A $1.5$ g sample containing oxalic acid and some inert impurity was dissolved in enough water and volume was made up to $250$ mL. A $20$ mL portion of this solution was then mixed with $30$ mL of an alkali solution. The resulting solution was then treated with stoichiometric amount of $C a C l_{2}$ just needed for precipitation of oxalate as $C a C_{2} O_{4}$ solution was filtered off and filtrate was finally titrated against $0.1 M H C l$ solution. $8.0$ mL of acid was required to reach the equivalence point. At last the above neutral solution was treated with excess of $A g N O_{3}$ solution and $A g C l$ obtained was washed, dried and weighed to be $0.4305$ g. Determine the mass percentage of oxalic acid in the original sample.

25

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67

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17.5

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82.5

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Solution

The correct option is C $82.5$
Molar mass AgCl is 143.3 g/mol.
0.4305 g of AgCl $= \frac{0.4305}{143.3} = 0.003$ moles. This is equal to moles of chloride ions.
Moles of chloride ions from HCl $= \frac{8.0 \times 0.1}{1000} = 0.0008$ moles
Moles of chloride ions from caclium chloride $= 0.003 - 0.0008 = 0.0022$ moles
Moles of calcium chloride $= \frac{0.0022}{2} = 0.0011$ . This is equal to moles of oxalic acid.
The molar mass of anhydrous oxalic acid is 90 g/mol.
Mass of oxalic acid $= 90 \times 0.0011 = 0.099 g$
However, this is present in 20 mL portion of solution.
Mass of oxalic acid present in 250 ml of solution $= 0.099 \times \frac{250}{20} = 1.2375 g$
Percent of oxalic acid in sample $= \frac{100 \times 1.2375}{1.5} = 82.5$ %

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