Determine the amount of urea (NH2CONH2) to be added in 1000 g of water to decrease its vapour pressure by 25%.
Determine the amount of urea (NH2CONH2) to be added in 1000 g of water to decrease its vapour pressure by 25%.
Correct Answer: 1111
Approach Solution - 1
Given :
Mole concept is used here, and the relationship is:
\( P_o - P_s = \frac{n}{N+n} \)
- \( P_o \): Initial vapor pressure
- \( P_s \): Vapor pressure of the solution
- \( N \): Number of moles of solvent
- \( n \): Number of moles of solute
Step-by-Step Calculation:
- The equation simplifies to: \[ 4n = N + n \]
- Rearranging, we get: \[ n = \frac{N}{3} \]
- \( N \), the number of moles of solvent (water), is calculated as: \[ N = \frac{1000}{18} \] \[ N = 55.56 \]
- Now, substituting \( N \) into the formula for \( n \): \[ n = \frac{55.56}{3} = 18.52 \]
- The weight of urea is calculated as: \[ \text{Weight of urea} = n \times \text{Molar mass of urea (60 g/mol)} \] \[ \text{Weight of urea} = 18.52 \times 60 = 1111.1 \, \text{g} \]
Conclusion:
The amount of urea is approximately 1111.1 g.
Approach Solution -2
\(\frac{P^0-P_s}{P_s}=\frac{n_{solute}}{n_{solvent}}\)
= \(\frac{\frac{x}{60}}{\frac{1000}{18}}=\frac{P^0-0.75P^0}{0.75P^0}\)
\(⇒x=\frac{10000}{9}=1111\ gm\)
So , the correct answer is 1111 gm
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
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- 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 ______.
Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.