MHT CET 2016 :: Chemistry :: General questions

• Chemistry - General questions

 The reaction  takes place in two steps as

(i)    $ NO_{2}Cl(g) $   $\underrightarrow{k_{1}}$   $ NO_{2}(g)+Cl(g)$

 (ii)  $NO_{2}Cl(g)+Cl(g)$   $\underrightarrow{k_{2}}$   $ NO_{2}(g)+Cl(g)$

 Identify the reacttion intermediate.

Answer:

Explanation:

 Cl(g) is the reaction  intermediate involved in the formation of $NO_{2}$ (g) and $Cl_{2}$ (g). Reaction intermediates are those which are formed within the reaction and has no presence in the products. Cl(g) is involved in both the reactions

$ NO_{2}Cl(g) $   $\underrightarrow{k_{1}}$   $ NO_{2}(g)+Cl(g)$

   $NO_{2}Cl(g)+Cl(g)$   $\underrightarrow{k_{2}}$   $ NO_{2}(g)+Cl(g)$

Which of the following compound when treated with dibenzyl cadmium yields benzyl methyl ketone ?

Answer:

Explanation:

Acetyl chloride, when treated with dibenzyl cadmium, yields benzyl methyl ketone

5.0 g of sodium hydroxide(molar mass 40 g $mol^{-1}$ ) is dissolved in little quantity of water and the solution is diluted up to  100mL. what is the molarity of the resulting solution?

Answer:

Explanation:

 Given, W (mass of solute) =5.0 g

 m (molar mass of solute)=40 g mol^-1

 Volume=100 mL

 Molarity  is given as,

 M= Moles of solute/ Volume of solution (in L)

M= W x 1000/  m x volume  of solution (in mL)

  =$\frac{5.0 g \times 1000}{40 \times 100}$

 =1.25 mol d$m^{-3}$

In the dry cell, what acts as a negative electrode?

Answer:

Explanation:

 In a dry cell, zinc acts as a negative electrode.

In this cell , the outer zinc container is the negatively charged terminal

At anode  (negative terminal)

$Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}$

At cathode (positive terminal)

$2MnO_{2}(s)+2e^{-}+2NH_{4}Cl(aq)\rightarrow Mn_{2}O_{3}(s)+2NH_{3}(aq)+H_{2}O(l)+2Cl^{-}$

 In face centred cubic unit cell, what is the volume occupied?

Answer:

Explanation:

 The volume occupied by the face centred cubic unit cell is $\frac{16}{3} \pi r^{3}$

 Packing fraction or volume occupied

 $Z_{eff}\times$   volume of c-atom = $\frac{Z_{eff}\times\frac{4}{3}\pi r^{3}}{a^{3}}$

 for fcc,   $Z_{eff}=4=\frac{4\times\frac{4}{3}\pi r^{3}}{a^{3}}$    (where , a=1)

 = $\frac{16}{3} \pi r^{3}$

• Chemistry - General questions