Volume of H₂O added = 175 ml
<h3>Further explanation</h3>
Given
100 gm of a 55% (M/M) and 20% (M/M) nitric acid solution
Required
waters added
Solution
starting solution
mass H₂O = 45%=45 g
%mass of H₂O in new solution = 100%-20%=80%
Can be formulated for %mass H₂O :
For water mass=volume(density = 1 g/ml)
So volume added = 175 ml
Answer:
3.1°C
Explanation:
Using freezing point depression expression:
ΔT = Kf×m×i
<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>
Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:
9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m
Replacing in freezing point depression formula:
ΔT = 5.12°cm⁻¹×0.472m×1
ΔT = 2.4°C
As freezing point of benzene is 5.5°C, the new freezing point of the solution is:
5.5°C - 2.4°C =
<h3>3.1°C</h3>
<em />
Because anymore water will breakdown the bonds of your Oh groups
I don't know if this is the answer you are looking for but it would be flat unless the player pushed the tuning slide in.
