Question

5 g of a sample of brass were dissolved in 1 litre conc. \[{H_2}S{O_4}\] ​20 mL of this solution were mixed with \[KI\] and liberated iodine required 20 mL of 0.0327 N hypo solution for titration. Calculate the amount of copper in the alloy.

Answer 1

Hint: The extraction of brass with concentrated \[{H_2}S{O_4}\]​, produces copper sulphate.
The reaction mechanism is as follows:
\[2\left[ {Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O} \right]\]
\[2CuS{O_4} + 4KI \to 2{K_2}S{O_4} + 2Cu{I_2}\]
\[2Cu{I_2} \to C{u_2}{I_2} + {I_2}\]
\[2N{a_2}{S_2}{O_3} + {I_2} \to 2NaI + N{a_2}{S_4}{O_6}\]
From the above mechanism it is clear that
2 mole \[Cu\]= 1 mole \[{I_2}\]= 2 mole hypo solution …… (i)
In order to calculate the number of moles of copper the formula used will be one showing relationship (number of moles $ \times $molecular mass= mass of element).

Complete step by step answer:
So According to the question;
If 20 mL of brass solution formed reacts with 20 mL of 0.0327 N hypo
Then 1000 mL of brass solution formed will react with 1000 mL of 0.0327 N hypo
Therefore, number of moles of hypo solution used can be calculated as follows
Number of moles = Mass​/Molecular mass of hypo solution (158)
The mass isn’t given in question but it can be calculated with the help of equivalent weight and normality,
From normality the mass is given as:
Mass = $E \times N \times V$
Where,
$E$= Equivalent weight
$N$= Normality
$V$= Volume

Therefore number of moles of hypo solution is given by

\[ = \dfrac{{E \times N \times V}}{{1000 \times 158}}\]

Where, $E$= 158, \[N\]= 0.0327 and \[V\] = 1000 mL

Therefore number of moles of hypo solution used = $\dfrac{{158 \times 0.03327 \times 1000}}{{1000 \times 158}} = 0.0327$

From equation (i) No. of moles of \[Cu\] = No. of moles of hypo = 0.0327 mole

Therefore Mass of copper in brass = number of moles $ \times $molecular mass

\[ \Rightarrow 0.0327 \times 63.5 = 2.07645\]

Mass of copper = 2.07645 gm
Therefore, % of copper in brass = $\dfrac{{2.07645}}{5} \times 100 = 41.529\% $

41.529% of copper is present in given 5 gm of brass.

Note: One mole is referred as equivalent to Avogadro’s number which is \[6.022 \times {10^{23}}\;\], an element can have different molar mass values depending upon the weight of \[6.022 \times {10^{23}}\;\] of elements atoms.