A solution containing 2.54 gram of a salt $K_{x} H_{y} {(C_{2} O_{4})}_{z} . n H_{2} O$ per litre. 10 ml of this salt solution required 30 ml of 0.01 N NaOH for complete neutralisation. Same quantity of salt solution was also found to require 40 ml of 0.01 N $K M n O_{4}$ solution for complete oxidation. Consider all the hydrogen atoms present in the salt are replaceable i.e y number of hydrogen atoms. Find out x,y,z and n (x,y,z are simplest possible integers) (K=39) Report your answer as ${\frac{x + y - z}{n}}$

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Detailed Solution

With NaOH valency factor of $K_{x} H_{y}$ ${(C_{2} O_{4})}_{z} . n H_{2} O$ is y
Mili equivalent of salt=mili equivalent of NaOH
$\frac{2.54}{M} \times 10 \times y = 0.01 \times 30$ …………………. (1)
M=molecular weight of salt.
With $K M n O_{4}$ valency factor $K_{x} H_{y} {(C_{2} O_{4})}_{z} . n H_{2} O$ is 2z
$\frac{2.54}{M} \times 10 \times 2 z = 0.01 \times 40$ ……………… (2)
$\frac{M . e q . 1}{M . e q . 2} = \frac{y}{2 z} = \frac{3}{4}$
$\frac{y}{z} = \frac{3}{2}$
Apply charge balancing
$K_{x} H_{y} {(C_{2} O_{4})}_{z} . n H_{2} O$
$x + y - 2 z = 0 (y = 3, z = 2)$
$x + 3 - 4 = 0$
$x = 1$
$K_{1} H_{3} {(C_{2} O_{4})}_{2} . n H_{2} O$
Calculate ‘M’ from eq.1
$M = 254$
$39 + 3 + 88 \times 2 + 18 n = 254$
$n = 2$
Hence x=1, y=3, z=2
Ans. $\frac{1 + 3 - 2}{2} = 1$