A solution is prepared by mixing 0.3355 moles of NaNO3 (84.995 g mol-1) with 235.0 g of water (18.015 g mol-1). Its density is 1
1 answer:
Answer:
Molality = 1.428m
Explanation:
Molality, m, is an unit of concentration defined as the ratio between moles of solute and kg of solvent:
Molality: Moles solute / kg solvent.
In the problem, water is the solvent (Is the compound in the higher quantity) whereas NaNO₃ is the solute.
Moles of solute: 0.3355moles NaNO₃.
Kg solvent:
Thus, molality of the solution is:
Molality = 0.3355 moles NaNO₃ / 0.2350kg
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Answer:
The degree of dissociation of acetic acid is 0.08448.
The pH of the solution is 3.72.
Explanation:
The
The value of the dissociation constant =
Initial concentration of the acetic acid = [HAc] =c = 0.00225
Degree of dissociation = α
Initially
c
At equilibrium ;
(c-cα) cα cα
The expression of dissociation constant is given as:
Solving for α:
α = 0.08448
The degree of dissociation of acetic acid is 0.08448.
The pH of the solution ;