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The size of a raw mango shrinks to a much smaller size when kept in a concentrated salt solution. Which one of the following process can explain this?
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The size of a raw mango shrinks to a much smaller size when kept in a concentrated salt solution. Which one of the following process can explain this?
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Kindly go through the solution
Osmosis.
An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules:
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An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules:
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The vapour pressure of solution will be less than vapour pressure of pure solvent, so some vapour molecules will get condensed to maintain new equilibrium.
The INCORRECT statement is:
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The INCORRECT statement is:
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Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute are together sealed in a container. Over time
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Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute are together sealed in a container. Over time
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Vapour pressure over solvent is greater than that over solution.
At , the vapour pressure of is of and that of acetone is of . A solution of in acetone has a total vapour pressure of of . The false statement among the following is
At , the vapour pressure of is of and that of acetone is of . A solution of in acetone has a total vapour pressure of of . The false statement among the following is
The above mixture will show positive deviation from Raoult's law.
The density of NaOH solution is 1.2 gm cm⁻³. The molality of this solution is _______ m. (Round off to the Nearest integer) [Use: Atomic masses: Na 23.0u O: 16.0 u H:1.0 u Density of H₂O: 1.0g cm⁻³]
The density of NaOH solution is 1.2 gm cm⁻³. The molality of this solution is _______ m. (Round off to the Nearest integer) [Use: Atomic masses: Na 23.0u O: 16.0 u H:1.0 u Density of H₂O: 1.0g cm⁻³]
Let volume of solution = x ml
So mass of solution = 1.2x
And mass of water = x gm
Mass of solute = 0.2x
Molality = (W_solute * 1000) / (M_solute * W_solvent) = (0.2x * 1000) / (40 * x) = 5 m
1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 × 10⁻³ bar. The molar mass of the biopolymer is _______ × 10⁴ g mol⁻¹. (Round off to the Nearest integer) [Use: R = 0.083 L bar mol⁻¹ K⁻¹]
1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 × 10⁻³ bar. The molar mass of the biopolymer is _______ × 10⁴ g mol⁻¹. (Round off to the Nearest integer) [Use: R = 0.083 L bar mol⁻¹ K⁻¹]
? = CRT
2.42 * 10? ³ = ( (1.46/M_polymer) / 0.1 ) * 0.083 * 300
M_polymer = (1.46 * 0.083 * 300) / (2.42 * 10? ³ * 0.1)
= 14.96 * 10? gm = 15 * 10? gm
When 3.00 g of substance 'X" is dissolved in 100 g of CCl₄, it raises the boiling point by 0.60K. The molar mass of the substance 'X" is __________g mol⁻¹. (nearest integer)
[Given Kb for CCl₄ is 5.0 K kg mol⁻¹]
When 3.00 g of substance 'X" is dissolved in 100 g of CCl₄, it raises the boiling point by 0.60K. The molar mass of the substance 'X" is __________g mol⁻¹. (nearest integer)
[Given Kb for CCl₄ is 5.0 K kg mol⁻¹]
ΔTb = 0.6 K
ΔTb = Kb × m = Kb × (wt. × 1000)/ (mol wt. × Wsolvent (gm)
0.6 = 5 × (3 × 1000)/ (mol wt. × 100)
∴ mol wt. = (15 × 10)/ (0.6) = 1500/6
= 250 g/mole