JEE 2013 :: Advanced 2013 :: Chemistry 2013

• Advanced 2013 - Physics 2013
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The reactions of Cl₂  gas with cold-dilute and hot -concentrated NaOH in water  give sodium salts of two (different)  oxoacids of chlorine, P and Q , respectively. The Cl₂ gas reacts with SO₂ gas , in the presence of charcoal, to give a product R. R reacts with white phosphorous to give a  compound s , on hydrolysis , s gives an oxoacid of phosphorous T.

P and Q respectively, are the sodium salts of 

Answer:

Explanation:

Plan 2NaOH +Cl₂  $\underrightarrow {cold}$    $NaCl+NaOCl+H_{2}O$

                                                                           P

$6NaOH+3Cl_{2}$  $\underrightarrow {hot}$   $5NaCl+NaClO_{3}+3H_{2}O$

                                                                                  Q

$HOCl$    $\underrightarrow {NaOH}$     $NaOCl$

hypochlorous                         P

acid

 $Cl_{2}+SO_{2}\rightarrow SO_{2}Cl_{2}$

                                           R

$10SO_{2}Cl_{2}+P_{4}\rightarrow 4PCl_{5}+10SO_{2}$

                                                  S

$PCl_{5}+4H_{2}O\rightarrow H_{3}PO_{4}+ 5HCl$

                                               T

 A fixed mass m of gas is subjected to the transformation of states from K to L to M to N  and back to K as shown in the figure.

 The pair of isochoric  processes among the transformation of states is 

Answer:

Explanation:

L → M at constant V → isochoric  N→K

 A fixed mass m of gas is subjected to the transformation of states from K to L to M to N  and back to K as shown in the figure.

The succeeding operations that enable  this transformation of states are

Answer:

Explanation:

 Plan By  Boyle's law at constant temperature   $p\propto\frac{1}{V}$

 By Charles's law at constant pressure

 $V\propto T$

 Process taking place at

 Constant temperature → isothermal

Constant preesure →  isobaric

Constant Volume →  Isochoric

Constant heat →  adiabatic

P and Q are isomers of dicarboxylic acid C₄H₄O₄. Both decolourize Br₂/ H₂O . On heating, P  forms the cyclic anhydride

Upon  treatment with dilute alkaline  KMnO₄ .p as well as Q could produce  one or more than one form S, T and U

  Compounds formed from P and Q are, repectively 

Answer:

Explanation:

Plan alkenes declourise Br₂ water

 cis-isomer $\underrightarrow{ dil.KMnO_{4}}$  Meso isomers by syn addition

 trans-isomer  $\underrightarrow{ dil.KMnO_{4}}$ d(+) and l(-) isomers by syn addition

 thus, reacemic mixture

 Formation of anhydride from  dicarboxylic acid indicates cis-isomer

 P and Q are isomers of dicarboxylic acids

 P,Q   $\underrightarrow{ Br_{2} water}$  decolourised

 P and Q have (C=C) bond 

P  $\underrightarrow{\triangle}$ anhydride

 Thus , P is cis-isomer

 T and U  (in 1:1 molar ratio)  form optically inactive (racemic mixture) due to external cancellation

An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and filtrate (Q). The precipitate  P was found to dissolve in hot water. The filtrate (Q) remained unchanged when treated with H₂ S in a dilute mineral acid medium. However, it gave a precipitate (R)   with H₂ S  in an ammonical medium.The precipitate R gave a coloured solution (S) when treated with H₂O₂ in an aqueous NaOH medium.

The coloured solution S contains

Answer:

Explanation:

Plan PbCl₂ is soluble in hot water. In ammonical medium , cations of group III and IV may be precipitated as hydroxide or sulphide  and dissolved in H₂O₂ due to oxidation

Mixture of two inorganic salts

S is  $Na_{2}CrO_{4}$

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013