JEE 2014 :: Advanced 2014 :: Chemistry 2014

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014

The total  number(s) of stable conformers with non zero dipole moment for the  following compound is(are)

Answer:

Explanation:

Plan. This product can be solved by using concept of conformational analysis of given organic compound. To solve the question draw  the stable conformational structures of organic compound  and determine  the net resultant dipole moment

 The confirmation of the given compound are as follows

 Stable conformer (with  $\mu\neq0$ )

These three have non-zero dipole moment due to the non-cancellation of all dipole  moment created by C-Cl  and C-Br bond

Consider the following list of reagents, acidified K₂Cr₂O₇, alkaline KMnO₄, CuSO₄,  H₂O₂, Cl₂, O3, FeCl₃. HNO₃ and Na₂S₂O₃. The total number of reagents that can  oxidise aqueous iodide to iodine is 

Answer:

Explanation:

Acidified   K₂Cr₂O₇, CuSO₄,  H₂O₂, Cl₂, O₃, FeCl₃. and  HNO₃  oxidise aq. iodide to iodine. Alkaline  KMnO₄   oxidise aq. iodide to IO₃^-.

  Na₂S₂O₃ is a strong reducing agent which on reaction with I₂  produces I^-

$N_{2}S_{2}O_{3}+I_{2}\rightarrow2I^{-}+N_{2}S_{4}O_{6}$

 Therefore, no reaction takes place between   Na₂S₂O₃  and iodide ion. Hence, the correct integer is (7)

 MX₂  dissociates into M²⁺ and X^-  ions in an aqueous solution, with a degree of dissociation ( $\alpha$) of 0.5. The ratio of the observed depression of freezing point of the aqueous solution  to the value of the depression of freezing point in the absence of ionic dissociation is 

Answer:

Explanation:

$MX_{2}\rightarrow M^{2+}+2X^{-}$

 van't Hoff factor for any salt can be calculated using equation

               $i=1+\alpha(n-1)$

 where ,n= number of constituent ions

 $\therefore$        $ i(MX_{2})=1+\alpha(3-1)$

                                  $= 1+2\alpha$

$\frac{(\triangle T_{f})observed}{(\triangle T_{f})teoretical}=i=1+2\alpha$

   $\therefore$        i=1+2 x0.5

                       i=2

A compound  H₂X with molar weight of 80g is dissolved in a solvent having density of 0.4g ML^-1. Assuming no change in volume upon dissolution, the molality  of a 3.2  molar solution is

Answer:

Explanation:

 Plane  This problem can be solved by using concept of conversion of molarity into molality.

Molarity=3.2 M

 Let  volume  of solution=1000mL=volume of solvent

 Mass of solvent =1000 x 0.4 =400g

 Since, molarity of solution is 3.2 molar

  $\therefore$       $ n_{solute}=3.2 mol$

 Molality (m)=  $\frac{3.2}{400/1000}=8$

 Hence, correct integer is (8)

 If the  value of Avogadro number is $6.023\times 10^{23}mol^{-1}$ and the value of Boltzmann constant is $1.380\times 10^{-23}JK^{-1}$, then the number of significant  digits in the calculated value of the universal  gas constant is 

Answer:

Explanation:

Plan. This problem can be solved by using the concept involved in calculation of significant figure.

 Universal gas constant R=kN_A

Where, k= Boltzman constant and N_A = Avogradro number

 $\therefore$    $ R=1.380\times 10^{-23}\times 6.023\times 10^{23}J/K-mol$

    = 8.31174  $\equiv 8.312$

 Since k and N_A both have four significant figures, so the value of R is also rounded off up to 4 significant figures.

  [When  number is rounded off, the   number of significant  figure is reduced, the last digit  is increased by  1 if the following  digits $\geq$ 5 and is left as such it following digits is $\leq$  4]

 Hence, correct integer is (4)

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014