The molar mass of water (\( \text{H}_2\text{O} \)) is 18 g/mol. Since we have 1 mole of water:
\[ \text{Mass of solvent} = 1 \, \text{mol} \times 18 \, \text{g/mol} = 18 \, \text{g}. \]
The total mass of the solution is the sum of the mass of the solute and the mass of the solvent:
\[ \text{Total mass} = \text{Mass of solute} + \text{Mass of solvent} = 2 \, \text{g} + 18 \, \text{g} = 20 \, \text{g}. \]
The mass percent of \( X \) is given by:
\[ \% \text{mass of } X = \frac{\text{Mass of } X}{\text{Total mass}} \times 100 = \frac{2 \, \text{g}}{20 \, \text{g}} \times 100 = 10\%. \]
The mass percent of \( X \) in the solution is 10%.
The molar mass of the water insoluble product formed from the fusion of chromite ore \(FeCr_2\text{O}_4\) with \(Na_2\text{CO}_3\) in presence of \(O_2\) is ....... g mol\(^{-1}\):
0.1 mole of compound S will weigh ...... g, (given the molar mass in g mol\(^{-1}\) C = 12, H = 1, O = 16)
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is:
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]