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.
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 : \(A_2 \rightleftharpoons 2A\)

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


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 : \(A_2 \rightleftharpoons 2A\)
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,