The total number of correct statements, regarding the nucleic acids is A. RNA is regarded as the reserve of genetic…

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Q jee_main_2025_02_april_morning Molar Mass Calculations

0.1~mol

of the following given antiviral compound (P) will weigh

\times 10⁻¹~g

Antiviral nucleoside structural layout diagram for Q48 — The diagram displays the full multi-cyclic skeletal structure of an antiviral nucleoside molecule featuring explicit iodine and fluorine atomic positions.

(Given : molar mass in

g \cdot mol⁻¹

H: 1, C: 12, N: 14, O: 16, F: 19, I: 127)

Numerical Answer. Answer: 372 to 372

Solution

Related Formula

Standard molecular weight calculation formula system:

Molar Mass = Σ (Atom Count \times Atomic Weight)

Mass = Moles \times Molar Mass

Core Logic

Let's tabulate and sum up the constituent atoms based on the verified structural formula layout :

Counting positions meticulously yields a molecular identity formula of $C₉H₁₀FIN₂O₄$ $\mathrm{C_9H_{10}FIN_2O_4}$.
Calculating the total molar mass:

Mass = (9 \times 12) + (10 \times 1) + 19 + 127 + (2 \times 14) + (4 \times 16)

Mass = 108 + 10 + 19 + 127 + 28 + 64 = 372~g/mol

Detailed skeletal atomic breakdown calculation diagram for Q48 — The diagram displays the full multi-cyclic skeletal structure of an antiviral nucleoside molecule featuring explicit iodine and fluorine atomic positions.

Step 1: Final Mass Scaling

Find the mass of $0.1~mol$ $0.1\mathrm{~mol}$ of the sample:

Weight = 0.1 \times 372 = 37.2~g = 372 \times 10⁻¹~g

Hence, the integer value for the blank box is 372.

Pattern Recognition

When counting atoms in nucleoside structures, remember that each vertex in the ribose sugar ring represents a fully saturated carbon component unless double bonds or heteroatoms are explicitly drawn.

Chapter Mix

Class 12 Chemistry: Biomolecules Class 11 Chemistry: Some Basic Concepts of Chemistry

Q37 jee_main_2025_02_april_morning Amino Acids Properties

Identify the correct statement among the following:

A. $(1) All naturally occurring amino acids except glycine contain one chiral centre.$
B. $(2) All naturally occurring amino acids are optically active.$
C. $(3) Glutamic acid is the only amino acid that contains a -COOH group at the side chain.$
D. $(4) Amino acid, cysteine easily undergo dimerization due to the presence of free SH group.$

Solution

Related Formula

Dimerization of thiol pairs yields a disulfide linkage via an oxidation mechanism:

2R-SH [-H₂] R-S-S-R

Core Logic

Let's review each choice item factually layout-by-row:

Option 1 is inaccurate: Isoleucine and threonine contain two distinct chiral centers.
Option 2 is inaccurate: Glycine lacks a asymmetric center, rendering it optically inactive.
Option 3 is inaccurate: Aspartic acid similarly contains a carboxylic acid functional group along its side chain.
Option 4 is correct: Cysteine contains a highly active free functional $-SH$ $-\mathrm{SH}$ group that easily dimerizes to form Cystine through a disulfide loop matrix.

Pattern Recognition

Disulfide bridges formed by cysteine play a crucial role in maintaining the tertiary and quaternary structural configurations of complex protein folds.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q28 jee_main_2025_03_april_evening Vitamins Classification

Fat soluble vitamins are: A. Vitamin

B₁

B. Vitamin

C

C. Vitamin

E

D. Vitamin

B₁₂

E. Vitamin

K

Choose the correct answer from the options given below :

A. C & D Only
B. A & B Only
C. B & C Only
D. C & E Only

Solution

Related Formula

Vitamins are classified into two categories based on solubility:

Water-soluble: Vitamin B-complex and Vitamin C.
Fat-soluble: Vitamins A, D, E, and K.

Core Logic

Categorize each given vitamin:

Vitamin $B₁$ $\mathrm{B}_1$ (Thiamine) $arrow$ $\rightarrow$ Water-soluble
Vitamin $C$ $\mathrm{C}$ (Ascorbic acid) $arrow$ $\rightarrow$ Water-soluble
Vitamin $E$ $\mathrm{E}$ (Tocopherol) $arrow$ $\rightarrow$ Fat-soluble
Vitamin $B₁₂$ $\mathrm{B}_{12}$ (Cobalamin) $arrow$ $\rightarrow$ Water-soluble
Vitamin $K$ $\mathrm{K}$ (Phylloquinone) $arrow$ $\rightarrow$ Fat-soluble

Step 1: Select correct options

Only C (Vitamin $E$ $\mathrm{E}$) and E (Vitamin $K$ $\mathrm{K}$) are fat-soluble. Thus, the correct option is (4).

Pattern Recognition

Use the mnemonic 'ADEK' to remember the fat-soluble vitamins. All other vitamins (mainly B-complex and C) are water-soluble.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q jee_main_2025_07_april_morning Hormones Structure

Thyroxine, the hormone has the structure given below:

Chemical skeletal structure of thyroxine hormone for Q47 — The structural formula of Thyroxine shows four iodine atoms attached to the aromatic benzene rings.

The percentage of iodine in thyroxine is ______ %. (nearest integer) (Given molar mass in

g mol⁻¹

of

C: 12

,

H: 1

,

O: 16

,

N: 14

,

I: 127

)

Numerical Answer. Answer: 65 to 65

Solution

Related Formula

% I = 4 \times MIMThyroxine \times 100

Core Logic

Let's first determine the molecular formula of Thyroxine from its structure shown in below :

Thyroxine molecular structure highlight for Q47 — The structural formula of Thyroxine shows four iodine atoms attached to the aromatic benzene rings.

Formula: $C₁₅H₁₁O₄NI₄$

Now, compute the molecular mass:

Carbon: $15 \times 12 = 180$ $15 \times 12 = 180$
Hydrogen: $11 \times 1 = 11$ $11 \times 1 = 11$
Oxygen: $4 \times 16 = 64$ $4 \times 16 = 64$
Nitrogen: $1 \times 14 = 14$ $1 \times 14 = 14$
Iodine: $4 \times 127 = 508$ $4 \times 127 = 508$

Total Molecular Mass = 180 + 11 + 64 + 14 + 508 = 777 g mol⁻¹

Now, calculate the percentage of Iodine:

% I = (508)/(777) \times 100 \approx 65.38 % \approx 65 %

Hence, the percentage of iodine in thyroxine is $65$ $65$.

Pattern Recognition

Thyroxine molecular mass ( $777$ $777$) is composed mostly of Iodine ( $508$ $508$), making up approximately $508/777 \approx 65%$ $508/777 \approx 65\%$ of its entire mass. Always carefully account for the four phenyl-bound iodines.

Evaluation Rubric / Model Answer

Accurate molecular weight summation leading to $65%$ $65\%$ nearest integer Iodine composition.

Chapter Mix

Class 12 Chemistry: Biomolecules Class 12 Chemistry: Organic Chemistry - Oxygen containing compounds

Q43 jee_main_2025_07_april_morning Hydrolysis of Sucrose

Given below are two statements: Statement I:

D-(+)-glucose + D-(+)-fructose -H₂O sucrose sucrose Hydrolysis D-(+)-glucose + D-(+)-fructose

Statement II: Invert sugar is formed during sucrose hydrolysis. In the light of given statements, choose the correct answer from the options given below:

A. $Both Statement I and Statement II are true.$
B. $Statement I is false but Statement II is true.$
C. $Statement I is true but Statement II is false.$
D. $Both Statement I and Statement II are false.$

Solution

Core Logic

Statement I: Sucrose is formed by condensation of $D-(+)-glucose$ $\text{D-(+)-glucose}$ and $D-(-)-fructose$ $\text{D-(-)-fructose}$ (levorotatory fructose, not dextrorotatory as claimed). Hydrolysis of sucrose yields $D-(+)-glucose$ $\text{D-(+)-glucose}$ and $D-(-)-fructose$ $\text{D-(-)-fructose}$. Thus, Statement I is false.

Statement II: Hydrolysis of dextrorotatory sucrose ( $+66.5°$ $+66.5^{\circ}$) yields a mixture of dextrorotatory glucose ( $+52.5°$ $+52.5^{\circ}$) and highly levorotatory fructose ( $-92.4°$ $-92.4^{\circ}$). Because the overall specific rotation of the mixture becomes levorotatory ( $-39.9°$ $-39.9^{\circ}$), the hydrolyzed mixture is called invert sugar. Thus, Statement II is true.

Pattern Recognition

Natural fructose is always levorotatory, $D-(-)-fructose$ $\text{D-(-)-fructose}$. Dextrorotatory fructose mentioned in Statement I is an immediate giveaway that the statement is false.

Chapter Mix

Class 12 Chemistry: Biomolecules

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Chapter Mix

Reference Study Guides

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Pattern Recognition

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Pattern Recognition

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