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Physisorption

1.0 g sample ...

Question

$1.0$ g sample of the Rochelle salt, $(N a K C_{4} H_{4} O_{6} \cdot 4 H_{2} O)$ $(M w = 282)$ on ignition, is converted into $N a K C O_{3}$ $(M w = 122)$ which is titrated with $50$ mL of $0.1 M H_{2} S O_{4}$ . The excess of $H_{2} S O_{4}$ requires $30$ mL $0.2 M K O H$ . The percentage purity of Rochelle salt is :

36.4

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37.4

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38.4

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Solution

The correct option is B $36.4$
The reaction is:
$N a K C_{4} H_{4} O_{6} 1 m o l .4 H_{2} O Δ \to N a K . C 1 m o l O_{3} + 3 C O_{2} ↑ + H_{2} O ↑$
Now,
Total $m E q$ of $H_{2} S O_{4} = 50 \times 0.1 \times 2$ $(n$ factor $) = 10.0$
Excess $m E q$ of $H_{2} S O_{4} = m E q$ of $N a O H$ used $= 30 \times 0.2 \times 1$ $(n$ factor $)$ $= 6.0$
$m E q$ of $H_{2} S O_{4}$ used for $N a K C O_{3}$ $= m E q$ of $H_{2} S O_{4} (t o t a l) - m E q$ of $H_{2} S O_{4}$ used for $N a O H$ $= 10 - 6 = 4.0$
$∴ m E q$ of $N a K C O_{3} = 4.0$
$\Rightarrow n -$ factor of $N a K C O_{3} = 2$ , during its neutralisation with $H_{2} S O_{4}$ $(E w$ of $N a K C O_{3} = \frac{122}{2})$
Weight of $N a K C O_{3} = 4.0 \times 10^{- 3} \times \frac{122}{2} = 0.244 g$
$1$ mol of Rochelle salt $\equiv 1$ mol of $N a K C O_{3}$
Therefore, weight of Rochelle salt $= \frac{282 \times 0.244}{122} = 0.364 g$ .
% Purity of Rochelle salt $= \frac{0.364 \times 100}{1.0} = 36.4$ %

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Similar questions

Q. On ignition, Rochelle salt,

N a K C_{4} H_{4} O_{6} .4 H_{2} O

(molar mass

= 282

g/mol) is converted into

N a K C O_{3}

(molar mass

= 122

g/mol).

0.9546

g sample of the Rochelle salt on ignition gives

N a K C O_{3}

, which is titrated with

41.72

H_{2} S O_{4}

. From the following data, find the percentage purity of the Rochelle salt. The solution after neutralisation required

1.91

mL of

0.1297 N N a O H

. The

H_{2} S O_{4}

used for the neutralisation requires its

10.27

mL against

10.35

mL of

0.1297 N N a O H

1

g of a sample of

N a O H

was dissolved in

50

mL of

0.33 M

alkaline solution of

K M n O_{4}

and refluxed till all the cyanide was converted into

O C N^{⊝}

. The reaction mixture was collected and its

5

mL portion was acidified by adding

H_{2} S O_{4}

in excess and then titrated to endpoint against

19.0

mL of

0.1 M

F e S O_{4}

solution. The percentage purity of

N a C N

sample is:

1

g sample of

A g N O_{3}

is dissolved in

50

mL of water. It is titrated with

50

mL of

K I

solution. The

A g I

precipitated is filtered off. Excess of

K l

in filtrate is titrated with

\frac{M}{10}

K I O_{3}

in presence of

6 M H C l

till all

I^{-}

converted into

I C l

. It requires

50

mL of

\frac{M}{10}

K I O_{3}

solution.

20

mL of the same stock solution of

K I

requires

30

mL of

\frac{M}{10}

K I O_{3}

A g N O_{3}

in sample. The reaction is given below:

K I O_{3} + 2 K I + 6 H C l ⟶ 3 I C l + 3 K C l + 3 H_{2} O

2.0

g sample of

N a C N

is dissolved in

50

mL of

0.3 M

mild alkaline

K M n O_{4}

and heated strongly to convert all the

C N^{⊝}

O C N^{⊝}

. The solution after acidification with

H_{2} S O_{4}

requires

500

mL of

0.05 M

F e S O_{4}

. The percentage purity of

N a C N

in the sample is :

638.0

g of

C u S O_{4}

solution is titrated with excess of

0.2 M

K I

solution. The liberated

I_{2}

required

400

mL of

1.0 M

N a_{2} S_{2} O_{3}

for complete reaction. The percentage purity of

C u S O_{4}

in the sample is:

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1.0 g sample of the Rochelle salt, (NaKC4H4O6⋅4H2O) (Mw=282) on ignition, is converted into NaKCO3 (Mw=122) which is titrated with 50 mL of 0.1MH2SO4. The excess of H2SO4 requires 30 mL 0.2MKOH. The percentage purity of Rochelle salt is :