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

Question 14 Marks

For the reaction $A \rightarrow$ products.

The concentration of A at 10 minutes is __________ $\times 10^{-3} mol L ^{-1}$ (nearest integer).
The reaction was started with $2.5 mol L ^{-1}$ of A.

Answer

(2435)
Explanation: $t _{1 / 2} \propto[A]_0 \Rightarrow$ Order $=$ zero
$t _{1 / 2}=\frac{ A _0}{2 K} \Rightarrow$ Slope $=\frac{1}{2 K}=76.92$
$K=\frac{1}{2 \times 76.92}$
$[ A ]_{10}=- Kt + A _0=-\frac{1}{2 \times 76.92} \times 10+2.5=2.435$
$=2435 \times 10^{-3} mol / L$

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Question 24 Marks

Consider the following equilibrium,
$CO(g)+2 H_2(g) \rightleftharpoons CH_3 OH(g)$
0.1 mol of CO along with a catalyst is present in a $2 dm ^3$ flask maintained at 500 K . Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of $CH _3 OH$ is formed. The $K _{ p }^0$ is __________ $\times 10^{-3}$ (nearest integer).
Given : $R =0.08 dm ^3$ bar $K ^{-1} mol^{-1}$
Assume only methanol is formed as the product and the system follows ideal gas behaviour.

Answer

(74)
Explanation: $CO ( g )+2 H _2(g) \rightleftharpoons CH _3 OH ( g )$
$\begin{array}{llll} t =0 & 0.1 mol & a mol & - \\ t _{ eq } & 0.1- x & a -2 x & x =0.04 \\ & =0.06 & = a -0.08 & \\ & & =0.23-0.08 & \\ & & =0.15 \text { mole } & \end{array}$
$\begin{array}{l} V =2 L \\
T =500 K \\
P _{\text {total }}=5 bar \end{array}$
$n _{\text {Total }}=0.25=\frac{1}{4} mol$.
$P_{\text {total }}=n_{\text {total }} \times \frac{RT}{V}$
$\Rightarrow 5=(0.06+a-0.08+0.04) \times \frac{0.08 \times 500}{2}$
$\Rightarrow 10=(0.02+a) \times 0.08 \times 500$
$\Rightarrow a =0.25-0.02=0.23 mol$.
$K _{ P }=\frac{ X _{ CH _3 OH }}{ X _{ CO } \times X _{ H _2}^2} \times \frac{1}{\left( P _{ T }\right)^2}=\frac{0.04}{0.06 \times(0.15)^2} \times\left[\frac{1 / 4}{5}\right]^2$
$=\frac{4}{6 \times(0.15)^2 \times 16} \times \frac{1}{25}$
$=\frac{100 \times 100}{24 \times 225 \times 25}=\frac{100 \times 100}{135000}$
$=0.074=74 \times 10^{-3}$

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Question 34 Marks

0.1 mol of the following given antiviral compound $( P )$ will weigh __________ $\times 10^{-1} g$

(Given: molar mass in g mol^-1 H: 1, C : 12, N : 14, O: 16, F: 19, I: 127)

Answer

(372)
Explanation:

Molar mass $=372 gm$
$\therefore 0.1$ mole has $=372 \times 10^{-1} gm$

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Question 44 Marks

Consider the following electrochemical cell at standard condition.
$\begin{array}{l}Au(s)\left|QH_2, Q\right| NH_4 X(0.01 M)| | Ag^{+}(1 M) \mid Ag(s) \\
E_{\text {cell }}=+0.4 V\end{array}$
The couple $QH _2 / Q$ represents quinhydrone electrode, the half cell reaction is given below

$\left[\right.$ Given : $E _{ Ag ^{+} / Ag }^{ o }=+0.8 V$ and $\left.\frac{2.303 RT }{ F }=0.06 V\right]$
The $pK _{ b }$ value of the ammonium halide salt $\left( NH _4 X \right)$ used here is __________. (nearest integer)

Answer

(6)
Explanation: $QH _2+2 Ag ^{+} \rightarrow 2 Ag + Q +2 H ^{+}$
$E = E ^{\circ}-\frac{0.06}{2} \log \left[ H ^{+}\right]^2$
$E = E ^{\circ}-0.06 \times \log \left[ H ^{+}\right]$
$pH =-\log \left( H ^{+}\right)=\frac{ E - E ^{ o }}{0.06}=\frac{0.4-0.1}{0.06}$
$=\frac{0.3}{0.06}=5$
$\begin{array}{l} pH + NH _4 X =7-\frac{1}{2} pK _{ b }-\frac{1}{2} \log C \\
5=7-\frac{1}{2} \times pK _{ b }-\frac{1}{2} \log \left(10^{-2}\right) \\ pK _{ b }=6 .\end{array}$

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Question 54 Marks

A transition metal (M) among $Mn , Cr , Co$ and Fe has the highest standard electrode potential $\left( M ^{3+} / M ^{2+}\right)$. It forms a metal complex of the type $\left[ M ( CN )_6\right]^4$. The number of electrons present in the $e _{ g }$ orbital of the complex is __________ .

Answer

(1)
Explanation:
Co has highest standard electrode potential $\left( M ^{+3} / M ^{+2}\right)$ among $Mn , Cr , Co , Fe$
$\therefore$ Complex is $\left[ Co ( CN )_6\right]^4$ and its splitting is as follows.

$\therefore$ electron in $e _{ g }$ orbital is one.

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JEE SECTION - B [CHEMISTY - NUMERIC]

SECTION - B [CHEMISTY - NUMERIC]

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