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1992 WAEC Chemistry Practical All your burette readings (initial and final), as well as the size of your pipette,...

Chemistry
WAEC 1992

All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experimental procedure is required. All calculations must be done in your answer book.

F is a solution, O a dibasic acid H\(_2\)X. G is a solution containing 1.00g of sodium hydroxide in 250cm\(^3\) of solution.

(a) Put F in the burette and titrate with 20cm\(^3\) or 25cm\(^3\) portion of G using methyl orange as indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of F used 

(b) From your results and the information provided, calculate the;

(i) concentration of G in mol. dm\(^3\)

(ii) concentration of F in mol. dm\(^{-3}\)

(iii) molar mass of the acid H\(_2\)X, given that 100cm\(^3\) of solution F contained 0.4850 the acid.

The equation for the reaction is H\(_2\)X\(_{(aq)}\) + 2NaOH\(_{(aq)}\) \(\to\) Na\(_2\)X\(_{(aq)}\) + 2H\(_2\)O [H = 1; O = 16; Na = 23] 

 

 

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Explanation

Volume of pipette (G) = 25.00cm\(^3\) 

(a) The average volume of F used = 24.80cm\(^3\) 

(b)(i) Molar mass of NaOH = 40g/mol

Given that solution G contains 1.00g of NaOH in 250cm\(^3\) of solution = 4.00g in 1000cm\(^3\) of solution 

molar concentration of G \(\frac{4}{40}\) = 0.100 mol. dm\(^{-3}\) 

(ii) The equation of the reaction is H\(_2\)CX\(_{(aq)}\) + 2NaOH\(_{(aq)}\) + 2H\(_2\)O

\(\frac{\text{concentration of F x volume of F}}{\text{concentration of G x volume of G}}\) = mole ratio of \(\frac{F}{G}\) 

\(\frac{C_1 \times 24.80}{0.100 \times 25.00}\) = \(\frac{1}{2}\) 

C\(_1\) = \(\frac{0.100 \times 25.00}{24.80 \times 2}\) 

C\(_1\) = 0.0504m

concentration of F = 0.504 mol dm\(^{-3}\)

(iii) Molar mass of H\(_2\)X\(_3\)

Given that 1000cm\(^3\) 1000 cm\(^3\) of solution of F \(\to\) 0.485 of acid 

1000cm\(^3\) of sol. of F \(\to\) 485 x \(\frac{100}{10}\) = 4.85g/dm\(^3\)

Molar mass of H\(_2\)X = \(\frac{\text{mass concentration}}{molarity}\) 

\(\frac{4.85 g/dm^3}{0.050mol/dm^{-3}}\) = 96 gmol\(^{-1}\)

 

 


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