Sugar would not increase the boiling point of the water as much as the others.To show this, we recall the equation for boiling point elevation ∆t:
∆t = i Kb m
The difference in these four solutions will be in the product of the van't Hoff Factor i and molality m: i * m.
If we assume that we have 0.1m of each solute, the products i*m for these solutions are:
Solution Identity of particles i i * m
0.1 m NaCl Na+, Cl- 2 2 * 0.1 = 0.2
0.1 m MgSO4 Mg^2+, SO4^2- 2 2 * 0.1 = 0.2
0.1 m K2SO4 two K+, SO4^2- 3 3 * 0.1 = 0.3
0.1 m C12H22O11 covalently bonded molecules 1 1 * 0.1 = 0.1
Therefore, having the lowest i * m, sucrose will have the lowest boiling point elevation.
To start off any questions, we write down the givens.
Mass CO2 = 168.2g
Molar Mass CO2 = <span>44.011g/mol
Moles CO2 = ?
To find the number of moles, we do mass/molar mass to find the # moles
n CO2 = 44.011g / 168.2 g/mol
n CO2 = </span>0.26165874 mol
n CO2 = 0.2617 mol
The number of moles in the sample is 0.2617 mol.
B. increase; decrease or are absent
In a liquid, the particles move around more than a solid, but way less than in a gas because there isn't that much space between particles.
When matter is a gas, the particles tend to move in a larger area and they tend to get very far away from each other. So, the space between particles increase and the attraction forces decrease.
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
=
(1)
and we also know that:
n = M x 
If we replace this expression in (1) we have:
x
=
x 
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is
:
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
Learn more about molarity here:
brainly.com/question/19517011
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