JEE 2015 :: Main 2015 :: Chemistry 2015

• Main 2015 - Physics 2015
• Main 2015 - Chemistry 2015
• Main 2015 - Mathematics 2015

Which of the following is the energy of a possible excited state of hydrogen?

Answer:

Explanation:

 Since, at n=1, the population of electrons.

 is maximum i.e, at ground state, So maximum

 excitation  will take  place from n=1 to n=2

 Hence, n=2 is the possible excited state.

 Now, we have the formula for energy of H-atom

  $\left(E_{n}\right)_{H} =-13.6 \frac{Z^{2}}{n^{2}}eV$

 where,   Z= atomic number, z for H-atom- 1.

     $\therefore$            $\left(E_{n}\right)_{H} =-13.6\times \frac{{1}}{2^{2}}eV$

                                   =  -3.4 eV

 Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately

Answer:

Explanation:

From this figure

      $(AC)^{2}=(AB)^{2}+(BC)^{2}$

     $(AC)^{2}=a^{2}+a^{2}=2a^{2}$

  Also,         $(AD)^{2}=(AC)^{2}+(DC)^{2}$

               $(4r)^{2}=(2a)^{2}+a^{2}$

      $16r^{2}=3a^{2}$

   $r=\frac{\sqrt{3}}{4}a$

Now, when Na metal crystallise in bcc unit cell with unit cell edge, a= 4.29Å

   We have the formula for radius

i.e,            $r=\frac{\sqrt{3}}{4}\times4.29$ Å  =1.86Å

 The molecular formula  of a commercial  resin used  for exchanging  ions in water softening is  $C_{8}H_{7}SO_{3}Na$    (mol.wt.=206) . What would be the maximum uptake of Ca²⁺ ions  by the resin  when expressed in mole per gram resin?

Answer:

Explanation:

 We know that the molecules weight of  $C_{8}H_{7}SO_{3}Na$ 

=  $12\times 8+1\times7+32+16\times 3+23=206$  We have to find  mole per gram  of resin.

  $\therefore$  1g of  $C_{8}H_{7}SO_{3}Na$   has number of mole

                     =   weight og given resin / molecular weight of resin

             = $\frac{1}{206}$  mol

 Now, reaction looks like

$2C_{8}H_{7}SO_{3}Na+Ca^{2+}\rightarrow (C_{8}H_{7}SO_{3})_{2}Ca+2Na^{+}$

 $\because$ 2 moles of  $C_{8}H_{7}SO_{3}Na$  combines with 1 mol Ca²⁺

 $\because$  1 mole of   $C_{8}H_{7}SO_{3}Na$ will combine  with   $\frac{1}{2}$ mol Ca²⁺

$\because$   $\frac{1}{206}$  mole of   $C_{8}H_{7}SO_{3}Na$  will combine with

$\frac{1}{2}\times\frac{1}{206}$ mol Ca²⁺   =   $\frac{1}{412}$  mol Ca²⁺

• Main 2015 - Physics 2015
• Main 2015 - Chemistry 2015
• Main 2015 - Mathematics 2015