<u>Answer:</u> The correct answer is Option d.
<u>Explanation:</u>
According to Lewis acid-base concept:
The substance which is donating electron pair is considered as Lewis base and the substance which is accepting electron pair is considered as Lewis acid.
For the given chemical reaction:
is accepting electron pair and is getting converted to 
. Thus, it is considered as Lewis acid.
present in CuO is a Lewis base because it is donating electron pair.
Thus, the correct answer is Option d.
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
Answer:
Explanation:
From the given information:
TO start with the molarity of the solution:
= 0.601 mol/kg
= 0.601 m
At the freezing point, the depression of the solution is 
Using the depression in freezing point, the molar depression constant of the solvent 
The freezing point of the solution 
The molality of the solution is:
Molar depression constant of solvent X, 
Hence, using the elevation in boiling point;
the Vant'Hoff factor 
NaClO, K2CO3, and NaF will form basic soltions.
