0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.
Show Hint
Correct Answer: 372
Approach Solution - 1
The problem asks for the weight of 0.1 mol of the given antiviral compound (P). To find this, we first need to calculate the molar mass of the compound from its chemical structure and the provided atomic masses.
Concept Used:
The relationship between mass, number of moles, and molar mass is given by the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar Mass} \]The molar mass of a compound is the sum of the atomic masses of all the atoms in its molecular formula.
Step-by-Step Solution:
Step 1: Determine the molecular formula of the compound (P) by counting the number of atoms of each element from its structure.
The compound consists of a 5-iodouracil base and a fluorinated deoxyribose sugar.
- Carbon (C): There are 4 carbon atoms in the pyrimidine ring and 5 carbon atoms in the sugar moiety. Total C atoms = 4 + 5 = 9.
- Hydrogen (H): There is 1 H on the pyrimidine ring (at C6), 1 H on a ring nitrogen (at N3), and 8 H atoms on the sugar moiety (at C1', C2', C3', C4', C5', and in the two -OH groups). Total H atoms = 1 + 1 + 8 = 10.
- Nitrogen (N): There are 2 nitrogen atoms in the pyrimidine ring.
- Oxygen (O): There are 2 oxygen atoms in the pyrimidine ring (as carbonyl groups) and 3 oxygen atoms in the sugar moiety (one in the furanose ring and two in -OH groups). Total O atoms = 2 + 3 = 5.
- Fluorine (F): There is 1 fluorine atom on the sugar moiety.
- Iodine (I): There is 1 iodine atom on the pyrimidine ring.
The molecular formula of compound (P) is \( \text{C}_9\text{H}_{10}\text{FIN}_2\text{O}_5 \).
Step 2: Calculate the molar mass of compound (P) using the given atomic masses.
Atomic masses (g/mol): H = 1, C = 12, N = 14, O = 16, F = 19, I = 127.
\[ \text{Molar Mass} = (9 \times \text{C}) + (10 \times \text{H}) + (1 \times \text{F}) + (1 \times \text{I}) + (2 \times \text{N}) + (5 \times \text{O}) \] \[ \text{Molar Mass} = (9 \times 12) + (10 \times 1) + (1 \times 19) + (1 \times 127) + (2 \times 14) + (5 \times 16) \] \[ \text{Molar Mass} = 108 + 10 + 19 + 127 + 28 + 80 \] \[ \text{Molar Mass} = 372 \, \text{g/mol} \]Step 3: Calculate the mass of 0.1 mol of compound (P).
\[ \text{Mass} = \text{Number of moles} \times \text{Molar Mass} \] \[ \text{Mass} = 0.1 \, \text{mol} \times 372 \, \text{g/mol} = 37.2 \, \text{g} \]Step 4: Express the result in the required format of \( \text{___} \times 10^{-1} \, \text{g} \).
We need to find a value \( x \) such that \( x \times 10^{-1} = 37.2 \).
\[ x = \frac{37.2}{10^{-1}} = 37.2 \times 10 = 372 \]So, the mass is \( 372 \times 10^{-1} \, \text{g} \).
The value to be filled in the blank is 372.
Approach Solution -2
Molar mass is given as 372 g/mol for compound (P).
Hence, for 0.1 mole, the mass will be: \[ \text{Mass} = \text{Molar mass} \times \text{Number of moles} = 372 \times 0.1 = 37.2 \, \text{g} \]
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Mole concept and Molar Masses Questions
- Number of moles of \(H^+\) required by \(1\) mole \(MnO_4^-\) to oxidize oxalate ion to \(CO_2\) is ____ .
- Calculate the Molarity of a Solution having density = \(1.5\) \(g/ml\). \(\%(w/w)\) of Solute is \(36\%\) and Molecular weight of Solute is \(36\) \(g/mol\)
- Calculate the Molarity of a Solution having density = \(1.5\) \(g/ml\). \(\%(w/w)\) of Solute is \(36\%\) and Molecular weight of Solute is \(36\) \(g/mol\)
- The mass of sodium acetate (CH3COONa) required to prepare 250 mL of 0.35 M aqueous solution is g. (Molar mass of CH3COONa is 82.02 g/mol)
If a substance ‘A’ dissolves in a solution of a mixture of ‘B’ and ‘C’ with their respective number of moles as \(n_a\), \(n_b\), and \(n_c\), the mole fraction of C in the solution is:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,