| To find the mole fraction of urea in an aqueous solution given the molality, we start by noting the essential relationships and calculations. |
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Here, the solute is urea (CH4N2O).
Given: Molality (m) = 4.44 m
Let the mass of water be 1 kg. This implies the number of moles of urea is 4.44 moles since molality is moles of solute per kg of solvent.
The mole fraction of a solute (urea) is given by:
\[ \text{Mole fraction of urea} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}} \]
Moles of water: Given the mass of water is 1 kg (1000 g) and the molar mass of water is 18 g/mol:
\[ \text{Moles of water} = \frac{1000}{18} \approx 55.56 \text{ moles} \]
Substitute these values into the equation:
\[ \text{Mole fraction of urea} = \frac{4.44}{4.44 + 55.56} \]
\[ \text{Mole fraction of urea} = \frac{4.44}{60} \approx 0.074 \]
To express the mole fraction in terms of \( x \times 10^{-3} \):
\[ 0.074 = x \times 10^{-3} \]
\[ x = 0.074 \times 10^{3} = 74 \]
| Therefore, the value of \( x \) is 74, which lies within the range (74 to 74). |
Molality ($m$) of urea is given as 4.44 $m$, meaning 4.44 moles of urea are dissolved in 1000 g of water.
Step 1: Mole fraction formula
\[ X_{\text{urea}} = \frac{\text{Moles of urea}}{\text{Moles of urea} + \text{Moles of water}} \]
Step 2: Calculate moles of water
\[ \text{Mass of water} = 1000 \, \text{g}, \quad \text{Molar mass of water} = 18 \, \text{g/mol}. \] \[ \text{Moles of water} = \frac{1000}{18} = 55.56. \]
Step 3: Substitute values into the mole fraction formula
\[ X_{\text{urea}} = \frac{4.44}{4.44 + 55.56}. \] \[ X_{\text{urea}} = \frac{4.44}{60.00} = 0.0740. \]
Step 4: Express mole fraction as $x \times 10^{-3}$
\[ X_{\text{urea}} = 74 \times 10^{-3}. \] \[ x = 74. \]
Final Answer: 74
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
| Sample | Van't Haff Factor |
|---|---|
| Sample - 1 (0.1 M) | \(i_1\) |
| Sample - 2 (0.01 M) | \(i_2\) |
| Sample - 3 (0.001 M) | \(i_2\) |
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 : \(A_2 \rightleftharpoons 2A\)