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

(Given : molar mass in mathrmg cdot mol^-1$\mathrm{g \cdot mol}^{-1}$ H: 1, C: 12, N: 14, O: 16, F: 19, I: 127)

### Related Formula \% mathrmI = frac4 times M_mathrmIM_textThyroxine times 100$$\% \mathrm{I} = \frac{4 \times M_{\mathrm{I}}}{M_{\text{Thyroxine}}} \times 100$$ ### Core Logic Let's first determine the molecular formula of Thyroxine from its structure shown in below :

- Formula: mathrmC_15mathrmH_11mathrmO_4mathrmNmathrmI_4$\mathrm{C}_{15}\mathrm{H}_{11}\mathrm{O}_4\mathrm{N}\mathrm{I}_4$ 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$4 \times 127 = 508$ textTotal Molecular Mass = 180 + 11 + 64 + 14 + 508 = 777 text g mol^-1$$\text{Total Molecular Mass} = 180 + 11 + 64 + 14 + 508 = 777 \text{ g mol}^{-1}$$ Now, calculate the percentage of Iodine: \% mathrmI = frac508777 times 100 approx 65.38 \% approx 65 \%$$\% \mathrm{I} = \frac{508}{777} \times 100 \approx 65.38 \% \approx 65 \%$$ 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

### Core Logic Statement I: Sucrose is formed by condensation of textD-(+)-glucose$\text{D-(+)-glucose}$ and textD-(-)-fructose$\text{D-(-)-fructose}$ (levorotatory fructose, not dextrorotatory as claimed). Hydrolysis of sucrose yields textD-(+)-glucose$\text{D-(+)-glucose}$ and textD-(-)-fructose$\text{D-(-)-fructose}$. Thus, Statement I is false. Statement II: Hydrolysis of dextrorotatory sucrose (+66.5^circ$+66.5^{\circ}$) yields a mixture of dextrorotatory glucose (+52.5^circ$+52.5^{\circ}$) and highly levorotatory fructose (-92.4^circ$-92.4^{\circ}$). Because the overall specific rotation of the mixture becomes levorotatory (-39.9^circ$-39.9^{\circ}$), the hydrolyzed mixture is called **invert sugar**. Thus, Statement II is true. ### Pattern Recognition Natural fructose is always levorotatory, textD-(-)-fructose$\text{D-(-)-fructose}$. Dextrorotatory fructose mentioned in Statement I is an immediate giveaway that the statement is false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Amino acids exhibit amphoteric behavior due to the simultaneously present basic amino (-textNH_2$-\text{NH}_2$) and acidic carboxyl (-textCOOH$-\text{COOH}$) functional groups: 1. **At pH = 2 (Highly Acidic Medium)**: The abundant concentration of hydronium ions (H^+$H^+$) protonates the carboxylate ion back into its non-ionized acid state (-textCOO^- rightarrow -textCOOH$-\text{COO}^- \rightarrow -\text{COOH}$), while the amine group remains securely protonated as an ammonium ion (-textNH_3^+$-\text{NH}_3^+$). Hence, product **A** exists purely as a **cation**. 2. **At pH = 10 (Highly Basic Medium)**: The high concentration of hydroxide ions (OH^-$OH^-$) abstracts protons from the system, deprotonating the carboxyl group into a carboxylate anion (-textCOO^-$-\text{COO}^-$) and neutralizing the ammonium group back into a free amine fraction (-textNH_2$-\text{NH}_2$). Hence, product **B** exists purely as an **anion**.

### Pattern Recognition Acidic environments (low pH) force positive overall charges onto amino structures (cation form). Basic environments (high pH) drive a net negative structure (anion form). This simple rule makes picking Option (1) immediate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Essential amino acids cannot be synthesized by the body and must be obtained from diet. From the given choices: * Valine (Essential) * Proline (Non-essential) * Lysine (Essential) * Threonine (Essential) * Tyrosine (Non-essential) Hence, (A), (C), and (D) are the essential amino acids. ### Pattern Recognition Mnemonic for essential amino acids: TV TILL PM MALL (Threonine, Valine, Tryptophan, Isoleucine, Leucine, Lysine, Phenylalanine, Methionine, Arginine, Histidine). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Statement I is false because glucose pentaacetate fixes the cyclic hemiacetal system structure securely into an unreactive ester configuration. As a result, it cannot revert to an open-chain form containing a free aldehyde group, meaning it does not react with carbonyl reagents like 2,4-DNP. Statement II is true because starch, a polysaccharide composed of glucose monomer blocks, undergoes acid-catalyzed hydrolysis to yield glucose when heated under pressure. ### Pattern Recognition Sees: Pentacetate reactivity vs polysaccharide hydrolysis. Shortcut: Acetylation locks the cyclic structure of glucose, preventing reactions that require an open-chain carbonyl group (like 2,4-DNP). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules