SECTION - B [CHEMISTY - NUMERIC] - JEE STD 11 Science Questions

Question 14 Marks

The hydrocarbon (X) with molar mass $80 g mol ^{-1}$ and $90 \%$ carbon has ___________ degree of unsaturation.

Answer

(3)
Sol. Mass of carbon $=\frac{80 \times 90}{100}=72 gm$
Number of C-atoms $=\frac{72}{12}=6$
Mass of hydrogen $=\frac{80 \times 10}{800}=8 gm$
Number of H -atoms $=\frac{8}{1}=8$
So molecular formula $C _6 H _8$
D.U. $=6+1-8 / 2=7-4=3$

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Question 24 Marks

The observed and normal masses of compound $MX _2$ are 65.6 and 164 respectively. The percent degree of ionisation of $MX _2$ is ____________%. (Nearest integer)

Answer

(75)
Sol.$\quad MX _2 \rightarrow M ^{+2}+2 X ^{-}$
$
\begin{array}{l}
i=\frac{\text { normal molar mass }}{\text { observed molar mass }} \\
i=\frac{164}{65.6} \\
1+(3-1) \alpha=\frac{164}{65.6} \\
2 \alpha=\frac{98.4}{65.6} \\
\alpha=0.75
\end{array}
$
percent dissociation $=75 \%$
Ans. 75

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Question 34 Marks

In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide $( AgBr )$. The percentage of Bromine in the organic compound is _________ $\times 10^{-1} \%$ (Nearest integer).
(Given : Molar mass of Ag is 108 and Br is $80 g mol ^{-1}$ )

Answer

(255)
Sol.
$\begin{aligned} \% \text { Bromine }= & \frac{\text { Molar Mass of Bromine }}{\text { Molar Mass of Silver bromide }} \\ & \times \frac{\text { Weight of AgBr }}{\text { Weight of sample }} \times 100 \\ = & \frac{80}{188} \times \frac{0.165}{0.25} \times 100 \\ = & \frac{4800}{188}=25.53=255 \times 10^{-1}\end{aligned}$

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Question 44 Marks

The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is ___________

Answer

(4)
Sol.

$n ($ stereogenic unit $)=2,2^2=4$ stereoisomers are possible.

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Question 54 Marks

Consider a complex reaction taking place in three steps with rate constants $k _1, k _2$ and $k _3$ respectively. The overall rate constant $k$ is given by the expression $k=\sqrt{\frac{k_1 k_3}{k_2}}$. If the activation energies of the three steps are 60,30 and $10 kJ mol ^{-1}$ respectively, then the overall energy of activation in $kJ mol ^{-1}$ is ___________ . (Nearest integer)

Answer

(20)
Sol. $K=\sqrt{\frac{ K _1 K_3}{K_2}}$
$
A \cdot e^{-E_2 / RT}=\sqrt{\frac{A_{e} e^{-E_1 / RT} \times A_3 e^{-Ea_3 / RT}}{A_2 e^{-E_2 / RT}}}
$
By comparinig exponential term
$
\begin{array}{l}
\frac{E_3}{R T}=\frac{1}{2} \times\left(\frac{E_{2_1}}{R T}+\frac{E_{2_3}}{R T}-\frac{E_{2_2}}{R T}\right) \\
E_a=\left(E_{a_1}+E_{a_3}-E_{a_2}\right) / 2 \\
E_a=(60+10-30) / 2=20 kJ mol^{-1}
\end{array}
$
Ans. 20

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JEE SECTION - B [CHEMISTY - NUMERIC]

SECTION - B [CHEMISTY - NUMERIC]

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