So, the answer would be 20 cm
Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
Explanation:
The reaction equation will be as follows.
Calculate the amount of 
dissolved as follows.
It is given that 
= 0.032 M/atm and 
= 
atm.
Hence, ![[CO_{2}]](https://tex.z-dn.net/?f=%5BCO_%7B2%7D%5D)
will be calculated as follows.
= 
= 
= 
or, = 
It is given that 
As, ![K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%7D)
![4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}](https://tex.z-dn.net/?f=4.46%20%5Ctimes%2010%5E%7B-7%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B0.608%20%5Ctimes%2010%5E%7B-5%7D%7D)
![[H^{+}]^{2}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%5E%7B2%7D)
= 
![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
= 
Since, we know that pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
So, pH = 
= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.
