SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 96

####34. An aqueous solution containing $0.10 g K I O_{3}$

$($ formula weight $= 214.0)$ was treated with an excess of KI solution. The solution was acidified with $H C l$ . The liberated $I_{2}$ consumed $45.0 m L$ of thiosulphate solution decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

$(1998, 5 M)$

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Solution:

The balanced equations are

(1) $2 M n C l_{2} + 5 K_{2} S_{2} O_{8} + 8 H_{2} O \to$

$2 K M n O_{4} + 4 K_{2} S O_{4} + 6 H_{2} S O_{4} + 4 H C l$

(2) $2 K M n O_{4} + 5 H_{2} C_{2} O_{4} + 3 H_{2} S O_{4} \to$

$K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 10 C O_{2}$

Given, mass of oxalic acid added $= 225 m g$

So, millimoles of oxalic acid added $= \frac{225}{90} = 2.5$

Now from equation 2

Millimoles of $K M n O_{4}$ used to react with oxalic acid=1 and Millimoles of $M n C l_{2}$ required initially $= 1$

$∴$ Mass of $M n C l_{2}$ required initially $= 1 \times (55 + 71) = 126 m g$

Alternative Method

$m$ moles of $M n C l_{2} = m$ moles of $K M n O_{4} = x$ (let)

and $M_{e q}$ of $K M n O_{4} = M_{e q}$ of oxalic acid

So, $x \times 5 = \frac{225}{90} \times 2$

Hence, $x = 1$

$∴ m$ moles of $M n C l_{2} = 1$

Hence mass of $M n C l_{2} = (55 + 71) \times 1 = 126 m g$ .

पिछला

अगला

Some Basic Concepts of Chemistry - Result Question 96

Solution:

Alternative Method

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Some Basic Concepts of Chemistry - Result Question 96

Solution:

Alternative Method

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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