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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

The equilibri...

Question

The equilibrium constant for the reaction in aqueous solution $H_{3} B O_{3} +$ glycerin $⇌$ complex, is 0.90. How many moles of glycerin should be added per litre of 0.10M $H_{3} B O_{3}$ so that 80% of the $H_{3} B O_{3}$ is converted to the boric-acid-glycerin complex?

4.44

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4.52

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3.6

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0.08

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Solution

The correct option is A 4.44

$H_{3} B O_{3}$ $+$
glycerine
$⟶$ complex
Initial
$0.10$
$0$
$0$
Equilibrium
$0.02$
$x$
$0.08$

The expression for the equilibrium constant is $K = \frac{[c o m p l e x]}{[g l y c e r i n e] [H_{3} B O_{3}]} = \frac{0.08}{x \times 0.02} = 0.9$
Thus, $x = 4.44$ .

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Similar questions

Q. The equilibrium constant for the reaction in aqueous solution

H_{3} B O_{3} (a q) + g l y c e r i n (a q) ⇌ (H_{3} B O_{3} - g l y c e r i n) (a q) i s 0.90

.
How many moles of glycerin should be added per litre of

0.10 M H_{3} B O_{3}

so that

80 %

of the

H_{3} B O_{3}

is converted to the boric acid -glycerin complex?

Q. The equilibrium constant for the reaction in aqueous solution

H_{3} B O_{3} (a q) + g l y c e r i n (a q) ⇌ (H_{3} B O_{3} - g l y c e r i n) (a q) i s 0.90

.
How many moles of glycerin should be added per litre of

0.10 M H_{3} B O_{3}

so that

80 %

of the

H_{3} B O_{3}

is converted to the boric -acid -glycerin complex?

Q. The equilibrium constant for the given reaction is

0.90

H_{3} B O_{3} + glycerine ⇌ H_{3} B O_{3} + glycerine complex

The amount of glycerine should be added to one litre of

0.10 M H_{3} B O_{3}

solution, so that 60% of the

H_{3} B O_{3}

is converted to boric acid-glycerine complex in

M

is :

Q. The equilibrium constant for the reaction in aqueous solution

H_{3} {B O}_{3} + g y l c e r i n ⇌ (H_{3} {B O}_{3} - g y l c e r i n)

0.90

.
How many moles of glycerin should be added per litre of

0.10 M

H_{3} {B O}_{3}

so that

80

% of the

H_{3} {B O}_{3}

is converted to the boric-acid-glycerin complex?

Q. Boric acid

(H_{3} B O_{3})

is:

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	$H_{3} B O_{3}$ $+$	glycerine	$⟶$ complex
Initial	$0.10$	$0$	$0$
Equilibrium	$0.02$	$x$	$0.08$

The equilibrium constant for the reaction in aqueous solution H3BO3+ glycerin ⇌ complex, is 0.90. How many moles of glycerin should be added per litre of 0.10M H3BO3 so that 80% of the H3BO3 is converted to the boric-acid-glycerin complex?

The correct option is A 4.44 H3BO3 + glycerine ⟶ complexInitial 0.10 0 0Equilibrium 0.02 x 0.08The expression for the equilibrium constant is K=[complex][glycerine][H3BO3]=0.08x×0.02=0.9Thus, x=4.44.

The equilibrium constant for the reaction in aqueous solution $H_{3} B O_{3} +$ glycerin $⇌$ complex, is 0.90. How many moles of glycerin should be added per litre of 0.10M $H_{3} B O_{3}$ so that 80% of the $H_{3} B O_{3}$ is converted to the boric-acid-glycerin complex?

The correct option is A 4.44

$H_{3} B O_{3}$ $+$
glycerine
$⟶$ complex
Initial
$0.10$
$0$
$0$
Equilibrium
$0.02$
$x$
$0.08$

The expression for the equilibrium constant is $K = \frac{[c o m p l e x]}{[g l y c e r i n e] [H_{3} B O_{3}]} = \frac{0.08}{x \times 0.02} = 0.9$
Thus, $x = 4.44$ .