The compound that would have the highest osmotic pressure when dissolved in water is 
.
So, option D is correct one.
The dissociation of one molecule of gives the maximum number of ions when dissolved in water ( 4 ions ) . Osmotic pressure is a colligative property and depends upon number of solute particles present in the solution . The solution having maximum number of solute particles will have maximum number of the osmotic pressure .
All other given molecules gives less number of number of ions when dissolved in water as compare to of .
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Answer:
There are many errors possible while titrating the acid of an unknown concentration with a base like NaOH.
Main error that leads to the error in results is misreading of the end point volume .
End point is when the reaction between the analyte and solution of known concentration has stopped .
Sometimes Burette is not straight enough to read the volume of the end point. One way to misread the volume of burette is by looking at the burette volume at an angle .
From above , volume seems to be higher. Indicators are used to indicate the color change of the reaction. In Acid-Base titrations , indicators first lighten up then changes its color.
So, error may have occurred in wrongly judging of the end point by color change of the indicator .
Answer:
Kr is a Noble Gas. Na is an alkali metal. F is halogen.
Group 17 is halogens. Inert is Noble Gases. Odourless and colourless is Noble Gases. Alkali metals do not occur freely in nature. Alkali metals are malleable
Explanation:
The molar mass of the unknown gas is 184.96 g/mol
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>How to determine the molar mass of the unknown gas </h3>
The following data were obtained from the question:
- Rate of unknown gas (R₁) = R
- Rate of CH₄ (R₂) = 3.4R
- Molar mass of CH₄ (M₂) = 16 g/mol
- Molar mass of unknown gas (M₁) =?
The molar mass of the unknown gas can be obtained as follow:
R₁/R₂ = √(M₂/M₁)
R / 3.4R = √(16 / M₁)
1 / 3.4 = √(16 / M₁)
Square both side
(1 / 3.4)² = 16 / M₁
Cross multiply
(1 / 3.4)² × M₁ = 16
Divide both side by (1 / 3.4)²
M₁ = 16 / (1 / 3.4)²
M₁ = 184.96 g/mol
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