JEE 2018 :: Chemistry (2018) :: Solved questions (2018)

• Chemistry (2018) - Solved questions (2018)

The solubility of salt of a weak acid (AB) at pH 3 is Y× 10^-3 mol L^-1. The value of Y is.........

[Given that the value of solubility product of AB (K_sp)=2× 10^-10 and the value of ionization constant of HB (K_a)=1× 10^-8]

Answer:

Explanation:

Let Solubility of AB in the buffer of pH 3=x

      $AB(s)\rightleftharpoons A^{+}(aq)+B^{-}(aq)$    K₁ =K_sp

     $B^{-}(aq)+H^{+}(aq)\rightleftharpoons HB^{}(aq)$   $K_{2}=\frac{1}{K_{a}}$

  $AB(s)+H^{+}(aq)\rightleftharpoons HB (aq)+A^{+}(aq)$

                                                           x              x

$K_{3}=\frac{K_{sp}}{k_{a}}$

 $K_{3}=\frac{[HB][A^{+}]}{[H^{+}]}=\frac{K_{sp}}{K_{a}}$

  $\therefore$  $\frac{x^{2}}{(10^{-3})}=\frac{2\times 10^{-10}}{1\times 10^{-8}}$

$\therefore$    x= 4.47 × 10^-3 M = Y× 10^-3 M

$\therefore$    y = 4.47

Liquids A  and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 torr. At the same temperature , a new solution of A  and B having mole fractions X_A and X_B , respectively , has vapour pressure of 22.5 torr. The value of X_A/X_B in the new solution is.........   (Given that the vapour pressure of pure liquid A is 20 Torr at temperature T )

Answer:

Explanation:

Use the formula

$p_{total}= p_A^0\times X_A +P_B^0\times X_{B}$

  and for equimolar solutions  $X_{A}=X_{B}=\frac{1}{2}$

Given , p_total =45 torr for equimolar solution 

        $p_A^0=20torr$

So,        $45= p_A^0\times \frac{1}{2}+p_B^0\times\frac{1}{2}=\frac{1}{2}(p_A^0+p_B^0)$

or     $ p_A^0+p_B^0=90torr$               .............(i)

      But we know $ p_A^0=20torr$

So  $ p_B^0$= 90-20=70 torr         (From Eq. (i))

  Now , for the new solution from the same formula

     p_total  = 22.5 torr

   22.5 = 20 X_A+70 (1-X_A)  [X_A + X_B=1]

    Or,         22.5 =70-50 X_A

   So,$X_{A}=\frac{70-22.5}{50}=0.95$

Thus, X_B =1-0.95 =0.05     (as X_A+ X_B =1)

Hence, the ratio

        $\frac{X_{A}}{X_{B}}=\frac{0.95}{0.05}=19$

A closed tank has two compartments  A and B , both filled with oxygen (assumed to be ideal gas) . The partition separating the two compartments is fixed and is a perfect heat insulator  (Fig .1) . If the old partition is replaced  by a new partition which  can slide and conduct  heat but does not allow the gas to leak across  (Fig. 2) , the volume ( in m³ ) of the compartment  A after the system attains equilibrium is................

Answer:

Explanation:

Given p₁=5 bar, V₁ =1 m³, T₁=400K

   So,          $n_{1}=\frac{5}{400R}$               (from pV=nRT)

          Similarly , p₂= 1 bar, V₂ = 3 m³, T₂ =300K

        $n_{2}=\frac{3}{300R}$

  Let  at equilibrium the new volume of A will be (1+x)

  So,  the new volume of B  will be (3-x) 

 Now, from the ideal gas equation

       $\frac{p_{1}V_{1}}{n_{1}RT_{1}}=\frac{P_{2}V_{2}}{n_{2}RT_{2}}$

     and at equilibrium (due to conduction of heat )

           $\frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}}$

 So     $\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}$ or V₁n₂ = V₂n₁

After putting the values

          $(1+x)\times \frac{3}{300R}=(3-x)\times \frac{5}{400R}$

 or       $(1+x)= \frac{(3-x)5}{4} or 4(1+x)= 15-5x$

  or $4+4x=15-5x$    or   $x=\frac{11}{9}$

 Hence, new volume of A i.e, (1+x) will comes as $1+\frac{11}{9}=\frac{20}{9} or 2.22$

For the electrochemical cell,

                Mg(s) | Mg ²⁺ (aq. 1M)|

                                           |Cu²⁺ (aq. 1M)| Cu(s)

 The standard emf of the cell is 2.70V at 300K. When the concentration of Mg ²⁺ is changed to x M, the cell potential changes to 2.67 V at 300K. The  value of x is .............. [ Given, $\frac{F}{R}=11500KV^{-1}$, where F is the Faraday constant and R is the gas contant, ln (10)=2.30)

Answer:

Explanation:

Equation  of cell reaction according  to the cell notation given, is

Given $E_{cell}^0=2.7V$. T= 300K

with [ Mg ²⁺ (aq)]= 1 M and [Cu ²⁺ (aq)]= 1 M and n=2

  Further . E_cell =2.67 V

  with [Cu ^2- (aq)]= 1M and [Mg ²⁺ (aq)]=xM

    and  $\frac{F}{R}=11500KV^{-1}$

   where F= Faraday constant,  R= gas constant

From the formula,

     $E_{cell}=E_{cell}^{0}-\frac{RT}{nF}ln\frac{[Mg^{2+}(aq)]}{[Cu^{2+}(aq)]}$

   After putting the given values

      $2.67= 2.70- \frac{RT}{2F}ln\frac{x}{1}$

or     $2.67= 2.70- \frac{RT}{2F}ln x$

           $-0.03=  \frac{-R\times 300}{2F}lnx$

Or      $ln x= \frac{0.003\times 2}{300}\times\frac{F}{R}$

             = $\frac{0.003\times 2 \times11500}{300}=2.30$

   So, ln x= 2.30

  or x=10 (as given  ln(10)=2.30)

Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose   unit cell is constructed from the unit cell of MX following the sequential instruction  given  below. Neglect the charge balance.

(a) Remove all the anions (X) except the central one.

 (b)  Replace all the face centered cations (M) by anions (X)

 (c)  Remove all the corner cations (M)

  (d)   Replace the central anion (X) with cation (M)

                The value of  ( Number of anions/ Number of cations )  in Z  is ..............

Answer:

Explanation:

The unit cell of initial structure of ionic solid MX looks like 

In NaCl type of solids cations (Na⁺ ) occupy the octahedral voids white anions (Cl^- ) occupy the face centre positions.

  However , as per the demand of problem the position of cations and anions are swapped. 

   We also know that (for 1 unit cell)

  (A) Total number of atoms at FCC= 4

   (B) Total number of octahedral voids =4'

( as no. of atoms at FCC= No. of octahedral voids)

Now taking the conditions one by one.

(i) If we remove all the anions except the central one than number of left anions.

                          =4-3=1

 (ii) If we replace all the face centred cations by anions than effective number of cations will be = 4-3=1

               Likewise  effective number of anions will be=1+3=4

   (iii) If we remove all the corner cations then effective number of cations will be 1-1=0

   (IV) If we replace central anion with cation then effective number of cations will be 0+1=1

       Likewise effective number of anions will be 4-1=3

         Thus, as the final outcome, total number of cations present  in Z after fulfilling  all the four sequential instructions=1

                  Likewise, total number of anions =3

      Hence, the value of

                 Number of anions/ Number of cations = $\frac{3}{1}=3$

• Chemistry (2018) - Solved questions (2018)