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:
Mass = 217.30 g
Solution:
Mass and moles are related to each other as,
Moles = Mass / M.mass ---- (1)
Where;
Moles = 3.42
M.Mass = 63.54 g.mol⁻¹
Solving eq. 1 for Mass,
Mass = Moles × M.mass
Putting Values,
Mass = 3.42 mol × 63.54 g.mol⁻¹
Mass = 217.30 g
When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
The correct answer is C.
Most of the time, double replacements produce one product that is soluble and one that is insoluble
This results in a precipitate within a liquid or aqueous solution
Hope this helps
Answer:
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