The ammonia <math> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </math> released on quantitative reaction of <math> <mn>0.6</mn> <mtext> </mtext> <mi mathvariant="normal">g</mi> </math> urea <math> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">O</mi> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </math> with sodium hydroxide <math> <mo>(</mo> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">O</mi> <mi mathvariant="normal">H</mi> <mo>)</mo> </math> can be neutralized by

Moles of H C l = moles of N H 3 <!- [endif]-><!- [endif]-> = 2 × moles of urea = 2 × 0.6 60 = 0.02 <!- [endif]-><!- [endif]-> ⇒ N . V = 0.02 <!- [endif]-><!- [endif]->

The ammonia <math> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </math> released on quantitative reaction of <math> <mn>0.6</mn> <mtext> </mtext> <mi mathvariant="normal">g</mi> </math> urea <math> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">O</mi> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </math> with sodium hydroxide <math> <mo>(</mo> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">O</mi> <mi mathvariant="normal">H</mi> <mo>)</mo> </math> &nbsp;can be neutralized by &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Ten Chemistry Class 11th Chemistry Some Basic Concepts of Chemistry

The ammonia $({N H}_{3})$ released on quantitative reaction of $0.6 g$ urea $({N H}_{2} {C O N H}_{2})$ with sodium hydroxide $(N a O H)$ can be neutralized by

Option 1 -

$100 m L$ of $0.1 N H C l$

Option 2 -

$100 m L$ of $0.2 N H C l$

Option 3 -

$200 m L$ of $0.2 N H C l$

Option 4 -

$200 m L$ of $0.4 N H C l$

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Moles of $H C l =$ moles of ${N H}_{3}$

$= 2 \times$ moles of urea $= 2 \times \frac{0.6}{60} = 0.02$

$\Rightarrow N . V = 0.02$

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The ammonia $({N H}_{3})$ released on quantitative reaction of $0.6 g$ urea $({N H}_{2} {C O N H}_{2})$ with sodium hydroxide $(N a O H)$ can be neutralized by

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The ammonia N H 3 released on quantitative reaction of 0.6 g urea N H 2 C O N H 2 with sodium hydroxide ( N a O H ) can be neutralized by

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The ammonia $({N H}_{3})$ released on quantitative reaction of $0.6 g$ urea $({N H}_{2} {C O N H}_{2})$ with sodium hydroxide $(N a O H)$ can be neutralized by