JEE 2014 :: Main 2014 :: Chemistry 2014

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
• Main 2014 - Mathematics 2014

 Among the following oxoacids, the correct decreasing  order of acid strength is 

Answer:

Explanation:

 Decreasing order of  strength of oxoacids 

    $ HClO_{4} > HClO_{3}> HClO_{2} >HOCl$

 The reason, Consider the structure of conjugate bases of each oxyacid of chlorine.

 The negative charge is more delocalised on ClO₄^-  due to resonance. Hence, ClO₄^- is more stable (and less basic).

Hence, we can say as the number of oxygen atoms around Cl-atom increases as oxidation number of Cl-atom increases and thus the ability of loose the H⁺ increases

 For the  non-stoichiometric reaction 2A+B→ C+D, the following kinetic data were obtained in three separate experiments, all at 298 K

  The rate law for the formation of C is 

Answer:

Explanation:

This problem can be solved by determining the order of reaction w.r.t  each reactant and  then writing rate law equation of the given equation accordingly as 

     $r=\frac{dC}{dt}=k[A]^{x}[B]^{y}$

  where x= order of reaction w.r.t A

        y= order of reaction w.r.t B

$1.2\times10^{-3}=k(0.1)^{x}(0.1)^{y}$

$1.2\times10^{-3}=k(0.1)^{x}(0.2)^{y}$

$2.4\times10^{-3}=k(0.2)^{x}(0.1)^{y}$

$R= k[A]^{1}[B]^{0}$

 As shown above, rate of reaction remains constant as the concentration of reactant (B)  changes from 0.1 M to 0.2 M  and becomes double when the concentration  of A change from 0.1 to 0.2  B(i.e doubled)

For the reaction ,

   $SO_{2}(g)+\frac{1}{2}O_{2}\rightleftarrows SO_{3}(g)$

  If K_P= K_C (RT)^x

 Where, the symbols have usual  meaning , then the value of x is (assuming ideality)

Answer:

Explanation:

For the given reaction, 

$\triangle n_{g}=n_{p}-n_{R}$ where n_P = number  of moles of products

 n_R= number of moles of reactants

  $K_{P}=K_{C}(RT)^{\triangle n_{g}}$

  $\triangle n_{g}=-(1/2)$

 Consider seperate  solution of .500M    $C_{2}H_{2}OH(aq)$  ,0.100M   $Mg_{3}(PO_{4})_{2}(aq),$  0.250 M KBr(aq)  and 0.125  M    $Na_{3}PO_{4}(aq)$  at 25° C. .Which statement is true about these solution, assuming all salts to be strong electrolytes?

Answer:

Explanation:

 This problem includes concept of colligative properties (osmotic pressure here)  and van't Hoff factor. Calculate  the effective molarity  to each solution

i.e, effective molarity=van't Hoff factor x molarity

 .500M    $C_{2}H_{2}OH(aq)$       i=1

 Effective molarity  =0.5

0.250 M KBr(aq)                                 i=2

 Effective molarity=0.5 M

0.100 M $Mg_{3}(PO_{4})_{2}(aq)$ i=5

Effective molarity=0.5 M 

0.125  M $Na_{3}PO_{4}(aq)$              i=4

Effective molarity=0.5 M

 Hence, all colligative properties are same

Note This equation is solved by assuming that the examiner has taken  $Mg_{3}(PO_{4})_{2}$  to be completely soluble. However, the fact is that it is insoluble (sparingly soluble)

 The equivalent  conductance of NaCl at concentration C and at infinite dilution are  $\lambda_{C}$ and   $\lambda_{\infty}$, respectively, The correct relationship between   $\lambda_{C}$ and   $\lambda_{\infty}$ is given as (where the constant B is positive)

Answer:

Explanation:

 According to Debye Huckel Onsager equation,

              $\lambda_{C}=\lambda_{\infty}-B\sqrt{C}$

 where,   $\lambda_{C}$ = limiting  equivalent conductivity at concentration C

$\lambda_{\infty}$  = limiting  equivalent  conductivity  at infinite dilution\

 C= concentration.

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
• Main 2014 - Mathematics 2014