3 Marks Question

Question 13 Marks

Silver is deposited on a metallic vessel by passing a current of 0.2 amps. for 3 hrs. Calculate the weight of silver deposited. (At mass of silver = 108 amu, 1 F = 96500 C)?

Answer

By Faraday's first law of electrolysis.
w = ZIt.
Substituting I = 0.2 A, t = 3×60×60 sec, Z= 108/(1×96500)
⇒ w = 2.417 g.
2.417 g of silver is deposited.

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Question 23 Marks

How the following conversions can be carried out?
i. 2-Bromopropane to 1-bromopropane
ii. Chloroethane to butane
iii. Benzene to diphenyl

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Question 33 Marks

How much copper is deposited on the cathode of an electrolytic cell if a current of $5$ ampere is passed through a solution of copper sulphate for $45$ minutes?

Answer

We know that
${\left[Cu=63.5 g\ mol^{-1}, 1 F=96500 C\ mol^{-1}\right]}$
$\ce{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}$
$m=Z \times I \times t$
$=\frac{63.5}{2 \times 96500} \times 5\ amp \times 45 \times 60$
$=\frac{857250}{193000}$
$=4.44 g$

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Question 43 Marks

An organic compound $A,$ which has a characteristic odour, on treatment with con. $NaOH$ forms two compounds $B$ and $C.$ Compound $B$ has molecular formula $C_7H_8O$ which on oxidation gives back $A.$ Compound $C$ is the sodium salt of an acid. $C,$ when heated with soda lime yields an aromatic hydrocarbon $D.$ Deduce the structures of $A, B, C$ and $D.$

Answer

This is Cannizzaro Reaction

The molecular formula of $(B)$ and characteristic odour of $(A)$ suggests that $(A)$ is an aromatic aldehyde, $C_6H_5CHO$ and $(B)$ is alcohol, $C_6H_5CH_2OH.$ As $(C)$ is a sodium salt of an acid gives hydrocarbon $(D)$ on heating with soda lime, $(C)$ is sodium benzoate and $(D)$ is benzene. In this reaction, Benzaldehyde undergoes self oxidation and reduction$($disproportionation$).$ Therefore:-

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Question 53 Marks

Give the major products that are formed by heating each of the following ethers with HI.

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Question 63 Marks

Write the mechanism of hydration of ethene to yield ethanol.

Answer

The mechanism of hydration of ethene to form ethanol involves three steps.

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Question 73 Marks

The first order rate constant for the decomposition of ethyl iodide by the reaction $C_2H_5I(g) \rightarrow C_2H_4(g) + HI(g)$
at $600 \ K$ is $1.60 \times 10^{-2} s^{-1}.$ Its energy of activation is $209 \ kJ/mol.$ Calculate the rate constant of the reaction at $700 \ Κ.$

Answer

Here $T_1=600 \ K$
$T_2=700 \ K$
$E_a=209 \ KJ / mol$
$=209000 \ Jmol^{-1}$
$k_1=1.60 \times 10^{-5} s^{-1}$
$k_2=?$
Using the formula
$\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]$
$\log k_2-\log k_1=\frac{E_a}{2.303 R}\left[\frac{700-600}{600 \times 700}\right]$
$\log k_2-\log 1.60 \times 10^{-5}=\frac{209000}{2.303 \times 8.314}\left[\frac{100}{600 \times 700}\right]$
$\log k_2=\log 1.60 \times 10^{-5}+2.599$
$\log k_2=-4.796+2.599$
$=-2.197$
$k_2=\text { anti } \log (-2.197)$
$=6.36 \times 10^{-3} s^{-1}$

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Question 83 Marks

Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Answer

A lead storage battery consists of a lead anode, a grid of lead packed with lead dioxide $(PbO_2)$ as cathode and a $38\%$ solution of sulphuric acid $(H_2SO_4)$ as an electrolyte.

When the battery is in use, the following cell reactions take place:
At anode: $Pb (s)+ SO _4^{2-}(a q) \rightarrow PbSO _4(s)+2 e^{-}$
At cathode: $PbSO _4(s)+ SO _4^{2-}(a q)+4 H ^{+}(a q)+2 e^{-} \rightarrow PbSO _4(s)+2 H _2 O (l)$
The overall cell reaction is given by,
$ Pb(s)+PbO_2(s)+2 H_2 SO_4(aq) \rightarrow 2 PbSO_4(s)+2 H_2 O(l) $
When battery is charging, the reverse of all these reactions takes place.
Hence, on charging, $PbSO_4(s)$ present at the anode and cathode is converted into $Pb$ and $PbO_2,$ respectively. Sulphuric acid $(H_2SO_4)$ reconstitute in the reaction. The reaction may written as:
$2 PbO _2(s)+2 H _2 O ( l ) \xrightarrow{\text { charging }} Pb ( s )+ PbO _2(s)+2 H _2 SO _4$

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