0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion, produced 0.9 g H₂O. Molar mass of (X) is ___________g mol\(^{-1}\).
To find the molar mass of the organic compound (X), we begin by determining the number of moles of hydrogen in 0.9 g of H₂O.
Step 1: Calculate moles of H₂O.
The molar mass of H₂O = 18 g/mol.
Number of moles = Mass / Molar mass = 0.9 g / 18 g/mol = 0.05 moles.
Step 2: Calculate moles of hydrogen in H₂O.
Each mole of H₂O contains 2 moles of hydrogen, so 0.05 moles of H₂O contain 0.1 moles of H.
Step 3: Relate moles of hydrogen to the compound (X).
The given organic compound (X) contains 10% hydrogen by mass. Thus, in 0.01 moles of (X), the moles of hydrogen = 0.1 moles (from combustion data).
Step 4: Calculate the molar mass of (X).
Since 10% of the molar mass is due to hydrogen: 0.1 M (where M = molar mass of X) = 0.01 moles of H, leading to M = 0.01 moles of H * 100 / 10% mol fraction of H = 0.1 moles / 0.01 * 100 = 100 g/mol.
Step 5: Verify the result.
The calculated molar mass of (X) = 100 g/mol, which is within the provided range of 100 to 100.
0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion, produced 0.9 g of H2O. Find the molar mass of compound (X).
Moles of compound \( X = 0.01 \, \text{mol} \)
Hydrogen content = 10% by mass
Mass of water produced = \( 0.9 \, \text{g} \)
Molar mass of water \( H_2O = 18 \, \text{g/mol} \)
\[ \text{Moles of } H_2O = \frac{0.9}{18} = 0.05 \, \text{mol} \]
Each mole of water contains 2 grams of hydrogen.
\[ \text{Mass of hydrogen} = 0.05 \times 2 = 0.1 \, \text{g} \]
If 0.1 g of hydrogen is 10% of the mass of 0.01 mol of X:
\[ \frac{0.1}{0.01 \times M} = 0.10 \]
Solving:
\[ 0.1 = \frac{0.1}{0.01M} \Rightarrow 0.01M = \frac{0.1}{0.1} = 1 \Rightarrow M = \frac{1}{0.01} = 100 \]
\[ \boxed{\text{Molar mass of } X = 100 \, \text{g/mol}} \]
Given that the compound contains 10% hydrogen, we can assume the molar mass of the compound is \( M_X \). - Mass of hydrogen in 0.01 mole of X = \( 0.01 \times 10 = 0.1 \text{g} \).
- In the complete combustion of X, the hydrogen reacts with oxygen to form H₂O.
- The number of moles of water formed is \( \frac{0.9}{18} = 0.05 \text{moles} \).
- In 1 mole of H₂O, there are 2 moles of hydrogen atoms. So, the moles of hydrogen atoms that reacted are \( 2 \times 0.05 = 0.1 \text{moles} \). The number of moles of hydrogen in 0.01 mole of X is 0.1 g, which gives the molar mass of X as: \[ \text{Molar Mass of X} = \frac{\text{Mass of X}}{\text{Number of moles of X}} = \frac{0.01}{0.01} = 100 \text{ g/mol}. \] Thus, the molar mass of X is 100 g/mol.





Consider the following reaction of benzene. the percentage of oxygen is _______ %. (Nearest integer) 
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]
A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_0\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _______. 