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

Question 14 Marks

$x~mg$ of $Mg ( OH )_2$ (molar mass $=58$) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is __________ mg. (Nearest integer)
(Given : $Mg ( OH )_2$ is assumed to dissociate completely in $H _2 O$)

Answer

3
$pH=10 $
$pOH=4 $
${\left[OH^{-}\right]=10^{-4}}$
no. of moles of $OH ^{-}=10^{-4}$
no. of moles of $Mg ( OH )_2=\frac{10^{-4}}{2}=5 \times 10^{-5}$
$\text {mass of } Mg ( OH )_2 =5 \times 10^{-5} \times 58 \times 10^3 mg $
$=2.9$

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Question 24 Marks

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185$ $S$ $cm ^2$ $mol^{-1}$ and the ionic conductance of hydroxyl and chloride ions are $170$ and $70$ $S$ $cm ^2$ $mol^{-}$ ${ }^1$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is $85.5$ $S$ $cm ^2$ $mol^{-1}$, its degree of dissociation is given by $x \times 10^{-1}$. The value of $x$ is ___________ . (Nearest integer)

Answer

3
$\lambda_{m}^{\circ} \text { of } NH_4 Cl=185 $
$\left(\lambda_{m}^{\circ}\right)_{NH_4^{+}}+\left(\lambda_{m}^{\circ}\right)_{Cl^{-}}=185 $
$\left(\lambda_{m}^{\circ}\right)_{NH_4^{+}}=185-70=115$ $S$ $cm^2$ $mol^{-1} $
$\left(\lambda_{m}^{\circ}\right)_{NH_4 OH}=\left(\lambda_{m}^{\circ}\right)_{NH_4^{+}}+\left(\lambda_{m}^{\circ}\right)_{OH^{-}} $
$=115+170 $
$\left(\lambda_{m}^{\circ}\right)_{NH_4 OH}=285 $ $S$ $cm^2$ $mol^{-1} $
degree of dissociation $=\frac{\left(\lambda_{ m }\right)_{ NH _4 OH }}{\left(\lambda_{ m }^{\circ}\right)_{ NH _4 OH }}$
$=\frac{85.5}{285} $
$=0.3 $
$=3 \times 10^{-1}$

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Question 34 Marks

A metal complex with a formula $MC \ell_4 \cdot 3 NH _3$ is involved in $sp ^3 d^2$ hybridisation. It upon reaction with excess of $AgNO _3$ solution gives 'x' moles of AgCl. Consider 'x' is equal to the number of lone pairs of electron present in central atom of $BrF _5$. Then the number of geometrical isomers exhibited by the complex is ____________ .

Answer

1 lone pair hence 1 mole AgCl.
Complex is $\left[ M \left( NH _3\right)_3 Cl _3\right] Cl$.
It shows 2 geometrical isomers $\left( Ma _3 b_3\right.$ type$)$ facial (fac) & meridional (Mer)

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Question 44 Marks

The amount of calcium oxide produced on heating 150 kg limestone ($75 \%$ pure) is ___________ kg. (Nearest integer)
Given : Molar mass (in $g~mol ^{-1}$) of Ca-40, O-16, C-12

Answer

63
$CaCO_3 \rightarrow CaO+CO_2 $
$\text { mass of } CaCO_3=\frac{150 \times 75}{100}=112.5 kg $
$=112500 g $
$n_{CaCO_3}=1125$
So moles of $CaO =1125$
$\text { mass of } CaO=\frac{1125 \times 56}{1000}=63 kg $
$\text { Correct answer } \Rightarrow 63$

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Question 54 Marks

Sea water, which can be considered as a 6 molar $(6 M )$ solution of NaCl, has a density of $2$ $g$ $mL ^{-1}$. The concentration of dissolved oxygen $\left( O _2\right)$ in sea water is 5.8 ppm. Then the concentration of dissolved oxygen $\left( O _2\right)$ in sea water, is $x \times 10^{-4} m$. $x =$ ________ . (Nearest integer)
Given: Molar mass of NaCl is $58.5$ $g$ $mol ^{-1}$
Molar mass of $O _2$ is $32$ $g$ $mol ^{-1}$

Answer

2
Sea water is 6 Molar in NaCl, So 1000 ml of sea water contains 6 mol of NaCl.
mass of solution $=$ Volume $\times$ density
$=1000 \times 2$
mass of solution $=2000 g $
$ppm =\frac{\text { mass of } O _2}{2000} \times 10^6 $
$\text { mass of } O _2=5.8 \times 2 \times 10^{-3}$
$=1.16 \times 10^{-2} g$
molality for $O _2=\frac{1.16 \times 10^{-2} / 32}{(2000-6 \times 58.5)} \times 1000$
$=\frac{1.16 \times 10}{32 \times 1649} $
$=0.000219 $
$=2.19 \times 10^{-4}$
Correct answer $\Rightarrow 2$

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JEE SECTION - B [CHEMISTY - NUMERIC]

SECTION - B [CHEMISTY - NUMERIC]

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