JEE 2015 :: Advanced 2015 :: Chemistry 2015

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015

When 100 mL of 1.0 M  HCl was mixed with 100mL   of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7° C was measured for the beaker and its contents (Expt. I). Because the enthalpy of neutralisation of a strong acid with a strong base is a constant  (-57.0kJmol-1), this experiment could be used to measure the calorimeter constant.In a second  experiment  (Expt.2), 100mL  of 2.0 M  acetic acid  (K_a = 2.0 x 10^-5) was mixed with 100 mL of 1.0 M   NaOH  (under identical conditions  to  Expt. 1)  where a temperature rise of 5.6° C was measured

The pH  of the solution after Expt.2 is 

Answer:

Explanation:

$CH_{3}COOH+NaOH \rightarrow CH_{3}COONa+H_{2}0$

200mol        100 mo                          0                   0

100mol          0                            100mmol          

A buffer is now formed.

                    $K_{a}=\frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=[H^{+}]$

                               $[\because [CH_{3}COOH]=[CH_{3}COO^{-}]]$

                             $\Rightarrow pH=pK_{a}=-\log (2\times 10^{-5})$

                         = 5- $\log 2$   = 4.7

When 100 mL of 1.0 M  HCl was mixed with 100mL   of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7° C was measured for the beaker and its contents (Expt. I). Because the enthalpy of neutralisation of a strong acid with a strong base is a constant  (-57.0kJmol-1), this experiment could be used to measure the calorimeter constant. In a second experiment  (Expt.2), 100mL  of 2.0 M  acetic acid  (K_a = 2.0 x 10^-5) was mixed with 100 mL of 1.0 M   NaOH  (under identical conditions to  Expt. 1)  where a temperature rise of 5.6° C was measured.

Enthalpy of disscociation (in KJmol^-1) of acetiic acid  obtained from the Expt. 2 is 

Answer:

Explanation:

Let C JK^-1 be the heat energy capacity of the calorimeter.

                        Maa of solution

             $=200 mL \times 1g mL^{-1}=200g $

 Heat evolved  in Exp.I

    =   $57\times 1000\times 0.1(mol)=5700J$

        $\Rightarrow  5700J=(200\times 4.2+C)\times 5.7$

                 $\Rightarrow  100=200\times 4.2+C$  ......(i)

Let x kJ/mol  is  heat evolved in neutralisation of acetic acid.

 $\Rightarrow x\times 1000\times 0.10=(200\times4.2+C)\times5.6$

             $\Rightarrow \frac{x\times 100}{5.6}=200\times4.2+C$         ...........(ii)

From (i)  and (ii): x=56 kJ/mol

    $\Rightarrow $       Enthalpy  of ionisation of acetic acid 

                                    = -56-(-57)= 1 kJ/MOL

The pair(s)  of ions where both the ions are precipitated upon passing H₂S  gas in presence of dilute HCl,  is (are)

Answer:

Explanation:

  Only radicals of I and II  group  of qualitative analysis get precipitated with H₂S  in the presence of dilute HCL

   (c)     $Cu^{2+}+H_{2}S$     $\underrightarrow{H^{+}}    $    $CuS \downarrow$ 

                                                                                                                    Black

          $Pb^{2+}+H_{2} S $       $\underrightarrow{H^{+}}  $    $PbS\downarrow $

                                                                                                                  Black

  (d)   $Hg^{2+}+H_{2} S $       $\underrightarrow{H^{+}}  $    $HgS\downarrow $

                                                                                                                  Black
            $Bi^{3+}+H_{2} S $       $\underrightarrow{H^{+}}  $    $Bi_{2}S_{3}$ $\downarrow $

                                                                       Brown ppt

            Ba²⁺ , Zn²⁺  and Fe³⁺  are not precipitated as sulphide.

The correct statement(s) regarding, $(i)  HClO,(ii) HClO_{2},(iii) HClO_{3}$  and  $HClO_{4}$  is(are)

Answer:

Explanation:

(a)number of Cl^-=O bonds in(ii) and (iii) together is three.Hence, wrong.

(b) number of  lone pair of on Cl in(ii) and (iii) together in three. Hence , correct

(c)In (iv) , Cl  is  $sp^{3}$  -hybridised. Hence, correct.

(d)amongest (i) and (iv) , the strongest acid is(iv). Hence , wrong

When O₂  is absorbed on a metallic surface, electron transfer occurs from the metal to O₂. The true statement (s)  regarding this adsorption is(are)  

Answer:

Explanation:

Since adsorption involves electron transfer from metal to O₂, it is chemical adsorption not physical adsorption at the (a) is incorrect. Aggregation is point focus which involves some bonding between adsorbent and adsorbate, hence exothermic. The last occupied molecular orbital in O₂  is   $\pi ^{*}2p$. Hence, electron transfer from metal to oxygen will increase the occupancy of$\pi ^{*}2p$    molecular orbitals. Also, increase in occupancy of $\pi ^{*}2p$ orbitals will decrease bond order and hence increase bond length of O₂.

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015