Chemistry NCERT Exemplar Solutions Class 11th Chapter Ten

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alok kumar singh

Contributor-Level 10

500 m L has H C l = 25 * 10 - 3 m o l = 25 m m o l

120 m L has = 1 m m o l

and 500 m L has C H 3 C O O H = 1 20 * 10 3 m m o l

so 20 m L has 10 3 20 * 20 500 = 2 m m o l

N a O H added in 20 m L is 5 * 1 2 = 2.5 m m o l

So, N a O H (left) + C H 3 C O O H ? C H 3 C O O H + H 2 O
1.5
2
1.5

p H = p K a + l o g ?   salt     acid   = 4.75 + l o g ? 0.4771 = 5.2271

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alok kumar singh

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The dotted ones are s p 2  carbons

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alok kumar singh

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30 m m o l L - 1 of H C l

Then in 250 m L H C l will be 30 4 m m o l

H 2 S O 4 will be 1 2 * 30 4 m m o l = 98 * 1 2 * 30 4 * 16 - 3 g = 0.3675 g

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alok kumar singh

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192.5

2 C ( g r ) + 3 H 2 ( g ) ? C 2 H 6 ( g )

Δ H f = 2 * Δ H c o m b ( C ) + 3 * Δ σ H c o m b H 2 - Δ H c o m b C 2 H 6

σ H f = 2 ( - 286 ) + 3 ( - 393.5 ) - ( - 1560 ) = 192.5

0.3675

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alok kumar singh

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N a C l + K 2 C r 2 O 7 ? H + C r O 2 C l 2 ? N a O H N a 2 C r O 4 ? H 2 O 2 C r O 5 ( C ) H +

  C r O 2 C l 2 5

N a 2 C r O 4 7

C r O 3 6

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alok kumar singh

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Vapour pressure over solvent is greater than that over solution.

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alok kumar singh

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Moles of H C l = moles of N H 3

= 2 * moles of urea = 2 * 0.6 60 = 0.02

N . V = 0.02

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alok kumar singh

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Kindly go through the solution

Conceptual.

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alok kumar singh

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Kindly go through the solution

 

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