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Plzz solve question no. 19
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Answer 1

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Q : One Gram of the acid C6H10O4 requires 0.768 g of KOH for complete neutralization.

How many neutralizable hydrogen atoms are in this molecule.

Final Answer :There are two neutralizable H atoms.

Formula Used :

1) No. of moles, = Given Mass /Molar Mass

2) No.of equivalents, = No. of moles * Valency Factor

Given :

1) Given Mass of C6H10O4 = 1 g

Molar Mass of C6H10O4 = 12*6+10*2+16*4=146g

Given Mass of KOH = 0.768 g

Molar Mass of KOH = 39+ 16 +1 = 56 g

Steps:

1) No. of acidic Hydrogen /neutralizable H atoms in acid be 'n'.

Valency Factor of Acid C6H10O4 = n

There is only one basic HYDROXIDE in KOH.

=> Valency Factor of KOH = 1

2) By Equivalent concept,

no. of equivalents of C6H10O4 = no. of equivalents of KOH

=> no. of moles * Valency Factor(C6H10O4)=

no. of moles of KOH

$= > n \times \frac{1}{146} =1 \times \frac{0.768}{56} \\ = > n = \frac{146 \times 0.768}{56} \\ = > n = 2 \: \: (approx.)$

Hence, No. of Neutralizable or unusable H in C6H10O4 are 2 .

Answer 2

Answer:

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