Question 14 Marks
Consider the following sequence of reactions :
Molar mass of the product formed (A) is _________ $g mol ^{-1}$.
Molar mass of the product formed (A) is _________ $g mol ^{-1}$.
Answer
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Sol.
Sol.
Questions
SECTION - B [CHEMISTY - NUMERIC]
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Question 24 Marks
$A \rightarrow B$
The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K . If the energy barrier with respect to reactant energy for such isomeric transformation is $191.48 kJ mol ^{-1}$ and the frequency factor is $10^{20}$, the time required for $50 \%$, molecules of A to become B is _________ picoseconds (nearest integer).
$
\left[R=8.314 J K^{-1} mol^{-1}\right]
$
The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K . If the energy barrier with respect to reactant energy for such isomeric transformation is $191.48 kJ mol ^{-1}$ and the frequency factor is $10^{20}$, the time required for $50 \%$, molecules of A to become B is _________ picoseconds (nearest integer).
$
\left[R=8.314 J K^{-1} mol^{-1}\right]
$
Answer
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$
\begin{array}{l}Sol.\
t_{1 / 2}=\frac{0.693}{K} \\
K=Ae^{-EaRT} \\
=10^{20} \times e^{-\frac{199.48 \times 10^3}{8.314 \times 1000}} \\
=10^{20} \times e^{-23.031}=10^{20} \times-e^{\ln 10 \times 10} \\
=\frac{10^{20}}{10^{10}}=10^{10} sec \\
t_{1 / 2}=\frac{0.693}{10^{10}}=6.93 \times 10^{-11} \\
=69.3 \times 10^{-12} sec .
\end{array}
$
$
\begin{array}{l}Sol.\
t_{1 / 2}=\frac{0.693}{K} \\
K=Ae^{-EaRT} \\
=10^{20} \times e^{-\frac{199.48 \times 10^3}{8.314 \times 1000}} \\
=10^{20} \times e^{-23.031}=10^{20} \times-e^{\ln 10 \times 10} \\
=\frac{10^{20}}{10^{10}}=10^{10} sec \\
t_{1 / 2}=\frac{0.693}{10^{10}}=6.93 \times 10^{-11} \\
=69.3 \times 10^{-12} sec .
\end{array}
$
Question 34 Marks
The number of molecules/ions that show linear geometry among the following is _________ .
$
SO_2, BeCl_2, CO_2, N_3^{-}, NO_2, F_2 O, XeF_2, NO_2^{+}, I_3^{-}, O_3$
$
SO_2, BeCl_2, CO_2, N_3^{-}, NO_2, F_2 O, XeF_2, NO_2^{+}, I_3^{-}, O_3$
Answer
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Sol. Linear species are$
Cl-Be-Cl, \quad O=C=O, \quad N^{-}=N^{+}=N^{-}
$
Sol. Linear species are$
Cl-Be-Cl, \quad O=C=O, \quad N^{-}=N^{+}=N^{-}
$
Question 44 Marks
In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl . The percentage composition of chlorine in the compound is _________ $\%$.
[Given : molar mass in $g mol ^{-1}$ of Ag : 108, $Cl =35.5]$
[Given : molar mass in $g mol ^{-1}$ of Ag : 108, $Cl =35.5]$
Answer
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$\begin{array}{ll}\text { Sol. } & n _{ Cl }= n _{ AgCl }=\frac{143.5 \times 10^{-3}}{143.5}=10^{-3} \\ & \% Cl =\frac{10^{-3} \times 35.5}{180 \times 10^{-3}} \times 100=19.72\end{array}$
$\begin{array}{ll}\text { Sol. } & n _{ Cl }= n _{ AgCl }=\frac{143.5 \times 10^{-3}}{143.5}=10^{-3} \\ & \% Cl =\frac{10^{-3} \times 35.5}{180 \times 10^{-3}} \times 100=19.72\end{array}$
Question 54 Marks
Some $CO _2$ gas was kept in a sealed container at a pressure of 1 atm and at 273 K . This entire amount of $CO _2$ gas was later passed through an aqueous solution of $Ca ( OH )_2$. The excess unreacted $Ca ( OH )_2$ was later neutralized with 0.1 M of 40 mL HCl . If the volume of the sealed container of $CO _2$ was x , then x is _________ $cm ^3$ (nearest integer).
[Given : The entire amount of $CO _2(g)$ reacted with exactly half the initial amount of $Ca ( OH )_2$ present in the aqueous solution.]
[Given : The entire amount of $CO _2(g)$ reacted with exactly half the initial amount of $Ca ( OH )_2$ present in the aqueous solution.]
Answer
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Sol. Let moles of $CO _2= n$
moles of $Ca ( OH )_2$ total initially $=2 n$
excess $Ca ( OH )_2= n$
gm equivalent of $Ca ( OH )_2=$ gm equivalent of HCl
$
\begin{array}{l}
n \times 2=0.1 \times \frac{40}{1000} \times 1 \\
n=2 \times 10^{-3}
\end{array}
$
Volume of $CO _2=2 \times 10^{-3} \times 22400=44.8 cm^3$
Sol. Let moles of $CO _2= n$
moles of $Ca ( OH )_2$ total initially $=2 n$
excess $Ca ( OH )_2= n$
gm equivalent of $Ca ( OH )_2=$ gm equivalent of HCl
$
\begin{array}{l}
n \times 2=0.1 \times \frac{40}{1000} \times 1 \\
n=2 \times 10^{-3}
\end{array}
$
Volume of $CO _2=2 \times 10^{-3} \times 22400=44.8 cm^3$