The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
<h3>What is the value of
van 't Hoff factor?</h3>
For most non-electrolytes dissolved in water, the Van 't Hoff factor is essentially $ 1 $ . For most ionic compounds dissolved in water, the Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
Which has highest Van t Hoff factor?
The Van't Hoff factor will be highest for
A. Sodium chloride.
B. Magnesium chloride.
C. Sodium phosphate.
D. Urea.
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The <span>covalent bonds are predicted for each atom are :
</span>(a)F = 1
(b) Si = 4
(c) Br = 1
(d) O = 2
(e) P = 3
(f) S = 2
Answer:
Exothermic
Explanation:
I think it is exothermic because heat is being released
Answer: Molar concentration of the tree sap have to be 0.783 M
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
where,
= osmotic pressure of the solution = 19.6 atm
i = Van't hoff factor = 1 (for non-electrolytes)
R = Gas constant = 
T = temperature of the solution = ![32^oC=[273+32]=305K](https://tex.z-dn.net/?f=32%5EoC%3D%5B273%2B32%5D%3D305K)
Putting values in above equation, we get:
Thus the molar concentration of the tree sap have to be 0.783 M to achieve this pressure on a day when the temperature is 32°C
Answer:
The correct option is: B. 33%
Explanation:
Orbital hybridisation refers to the mixing of atomic orbitals of the atoms in order to form new hybrid orbitals. The concept of orbital hybridization is used to explain the structure of a molecule.
The sp² hybrid orbitals are formed by the hybridization of one 2s orbital and two 2p orbitals. <u>The three sp² hybrid orbitals formed have 33% s character and 67% p character.</u>
