Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is x × 10–3. Value of x is _______. (integer answer)
Correct Answer: 74
Approach Solution - 1
| 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). |
Approach Solution -2
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
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 Solutions Questions
- What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg?
(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is 54.2 mm Hg at room temperature. Molar mass of glucose is 180 g mol–1 . - Solubility of \(Ca_3(PO_4)_2\) in \(100\) \(mL\) of pure water is \(W\) gm. Find out \(K_{sp}\) of \(Ca_3(PO_4)_2\) is: (M: Molecular mass of \(Ca_3(PO_4)_2\))
- We are given with three NaCl samples and their Van't Hoff factors. Which is correct option ?
Sample Van't Haff Factor Sample - 1 (0.1 M) \(i_1\) Sample - 2 (0.01 M) \(i_2\) Sample - 3 (0.001 M) \(i_2\) - Ethylene glycol of \(x\) \(kg\) is mixed with \(18.6\) \(kg\) of solvent, \(24 °C\) depression in freezing point takes place. Calculate value of \(x\). (Given: \(K_1\) = \(1.6 °C/molal\) M.W. of ethylene glycol = \(62\) \(g/mol\))
- $\text{Volume of } 3 \, \text{M NaOH (formula weight 40 g mol}^{-1}\text{) which can be prepared from } 84 \, \text{g of NaOH is } \_\_\_\_\_ \times 10^{-1} \, \text{dm}^3.$
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,