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

Question 14 Marks

The total number of hydrogen bonds of a DNA-double Helix strand whose one strand has the following sequence of bases is _____________.
$5^{'}$ - G - G-C-A-A-A-T-C-G-G-C-T-A-3'

Answer

Two nucleic acid chains are wound about each other and held together by H bonds between pair of bases.
Adenine from two hydrogen bonds with thymine and Guanine form three hydrogen bond with cytosine.
$5^{\prime} \mathrm{G}-\mathrm{G}-\mathrm{C}-\mathrm{A}-\mathrm{A}-\mathrm{A}-\mathrm{T}-\mathrm{C}-\mathrm{G}-\mathrm{G}-\mathrm{C}-\mathrm{T}-\mathrm{A}-3^{\prime}$
In given DNA strand total seven guanine and cytosine bases which form total 21 H-bonds and six adenine and thymine base which will form total 12 H-bonds with other DNA strand.
Total no. of H bonds $=7 \times 3+6 \times 2=33$
Ans. 33

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Question 24 Marks

The pH of a 0.01 M weak acid $\mathrm{HX}\left(\mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}\right)$ is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as $x \times 10^{-4} \mathrm{M}$. The value of x is _____________(nearest integer)

Answer

$\mathrm{HX}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \quad \mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}$
$0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad$ Not justified
$\Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3}$
$K_{a}=0.01 \alpha^{2}=10^{-8}$
On dilution let final concentration of $\mathrm{HX}=\mathrm{c} \mathrm{M}$
$Hx _{( aq )} \rightleftharpoons H _{( aq )}^{+}+ X _{( aq )}^{-}$
$C (1-\alpha) \quad C \alpha \quad C \alpha$
$\Rightarrow C \alpha=10^{-6}\qquad\ldots(1)$
$\frac{C \alpha^2}{1-\alpha}=K_a=10^{-8}\qquad\ldots(2)$
$\Rightarrow \frac{10^{-6} \alpha}{1-\alpha}=10^{-8}$
Data given is inconsistent & contradictory. This should be bonus.

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Question 34 Marks

Fortification of food with iron is done using $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_{4}.7 \mathrm{H}_{2} \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is _____________ (Nearest integer)
[Given : Molar mass of $\mathrm{Fe}, \mathrm{S}$ and O respectively are $56,32$ and $16 \mathrm{~g} \mathrm{~mol}^{-1}$]

Answer

Let mass of iron $=\mathrm{wg}$
$\Rightarrow \frac{\mathrm{W}}{150 \times 10^{3}} \times 10^{6}=12$
$\Rightarrow \mathrm{w}=150 \times 12 \times 10^{-3}=1.8 \mathrm{gm}$
Let mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}=\mathrm{w}_{1} \mathrm{gm}$
$\Rightarrow$ Moles of $\mathrm{Fe}=\frac{1.8}{56}=\left(\frac{\mathrm{w}_{1}}{56+96+7 \times 18}\right)$

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Question 44 Marks

$\mathrm{KMnO}_{4}$ acts as an oxidising agent in acidic medium. '$X$' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of '$d$' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of $\mathrm{X}+\mathrm{Y}$ is _____________.

Answer

$\underset{(\mathrm{O}. \mathrm{A})}{\stackrel{+7}{\mathrm{KMnO_{4}}}} \xrightarrow{\text { Acidic medium }} \mathrm{Mn}^{2+}$
X is difference in oxidation state.
$7-2=5$
So $\mathrm{X}=5$
$6 \mathrm{CH}_{3} \mathrm{COO}^{\ominus}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}$
$\rightarrow\left[\mathrm{Fe}_{3}\left(\mathrm{OH}_{2}\right)\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{6}\right]^{\oplus}+2 \mathrm{H}^{\oplus}$
$\left[\mathrm{Fe}_{3}(\mathrm{OH})_{2}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{6}\right]^{\oplus}+4 \mathrm{H}_{2} \mathrm{O}$
$\rightarrow \underset{\text { Brown red ppt }}{\left[\mathrm{Fe}(\mathrm{OH})_{2}\left(\mathrm{CH}_{3} \mathrm{COO}\right]\right.}+\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}^{\oplus}$
$\mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{d}^{5}~ 4 \mathrm{s}^{0}$ contains 5 d electrons
So $\mathrm{Y}=5$
$X+Y=5+5=10$

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Question 54 Marks

In Dumas' method for estimation of nitrogen 1g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _____________ $\%$ (nearest integer).
(Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$)

Answer

Partial pressure of $\mathrm{N}_{2}=(900-15)=885 \mathrm{~mm} \mathrm{~Hg}$
Mole of $\mathrm{N}_{2}=\frac{\left(\frac{885}{760} \times 0.15\right)}{(0.0821 \times 300)}=0.0071$ moles
$\%$ of nitrogen in organic compound
$=\frac{(0.0071 \times 28)}{1} \times 10$
$=19.85 \%$

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JEE SECTION - B [CHEMISTY - NUMERIC]

SECTION - B [CHEMISTY - NUMERIC]

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