MHT CET 2019 :: Chemistry :: General questions

• Chemistry - General questions

Which hydride among the following is the strongest reducing agent?

Answer:

Explanation:

Among the given hydrides of group 15 elements, BiH₃  is the strongest reducing agent. This is because as we move down the group, bond dissociation enthalpy decreases and thus reducing character increases.

 Hence , the order  is 

  NH₃  < PH₃ < AsH₃  < SbH₃  < BiH₃

 If the van't Hoff-factor for 0.1 M Ba(NO₃)₂  solution is 2.74, the degree of dissociation is 

Answer:

Explanation:

 Given, 

     Molarity =0.1 M

 vant Hoff factor (i)=2.74

 Since, i >1, it means  solute is undergoing dissociation.

$Ba(NO_{3})_{2}\rightleftharpoons Ba^{2+}+2NO_{3}^{-}$

 Number of particles dissociated (n)=3

 Now, $\alpha$   ( degree of dissociation )= $\frac{(i-1)}{(n-1)}$

                                                              = $\frac{(2.74-1)}{(3-1)}=0.87$

The resistance of $\frac {1}{10}$ M solution is 2.5 x 10³ ohm . What is the molar conductivity of solution ? (cell constant =1.25 cm^-1)

Answer:

Explanation:

 Given, Molarity = $\frac {1}{10}$ M

               R= 2.5 x 10³ $\Omega$

 Cell constant   $(\frac{l}{a})=1.25 cm^{-1}$

 Now, k( conductivity)=   $\frac{1}{k}\times\frac{l}{a}$

 $\therefore$    $k=\frac{1}{2.5\times 10^{3}}\times 1.25=0.5\times 10^{-3}\Omega^{-1}cm^{-1}$

 $ \Rightarrow$  Molar conductivity $(\wedge_{m})=\frac{k\times1000}{M}$

 $\therefore$       $\wedge_{m}=\frac{0.5\times 10^{-3}\times1000}{1/10}$

     $\wedge_{m}=50\Omega^{-1}cm^{2}mol^{-1}$

$C_{6}H_{5}COCH_{3}\xrightarrow[{Zn-Hg/conc.HCl}]{[H]}X_{,}X$   is

Answer:

Explanation:

$C_{6}H_{5}COCH_{3}\xrightarrow[{Zn-Hg/conc.HCl}]{4[H]} C_{6}H_{5}CH_{2}CH_{3}$

     Acetophenone                                            ethyl benzene

 The above reaction is known as Clemmensen reduction

Two electrolytic cells are connected in series containing  CuSO₄ solution and molten AlCl₃ . If in electrolysis  0.4 moles of 'Cu'  are deposited on the cathode of the first cell. The number of moles of 'Al'  deposited on the cathode of the second cell is 

Answer:

Explanation:

 Key Idea  First find the weight of copper deposited, by using  the formula number of moles=  Weight/ molecular weight

and then calculate the weight of Al deposited and the number of moles of Al by using the second law of Faraday's.

 Given,

 Number of moles of Cu deposited =0.4 moles

 According to Faraday's second law,

     weight of Cu  deposited / weight of Al deposited= Eq. wt. of Cu/ Eq. wt. of Al

 $\because$   No. of moles = $\frac {weight}{molecular weight}$

 $\therefore$    Weight of Cu = 0.4 x 63.5

 Now, from Eq.(i),

 = 0.4 x 6.5 / weight of Al deposited  = $\frac{\frac{63.5}{2}}{\frac{27}{3}}$

 $\therefore$  Weight of Al deposited=   $\frac{0.4 \times 63.5\times9}{31.75}=7.2 g$

 Now, number of moles of Al deposited = $\frac{7.2}{27}=0.27 $   moles

• Chemistry - General questions