(1a) An aqueous solution is made by dissolving 26.7 grams of sodium bromide in 4
ID: 487197 • Letter: #
Question
(1a) An aqueous solution is made by dissolving 26.7 grams of sodium bromide in 427 grams of water.
The molality of sodium bromide in the solution is= m.
(b)In the laboratory you are asked to make a 0.704 m manganese(II) iodide solution using 475 grams of water.
How many grams of manganese(II) iodide should you add? =
grams.
(c)In the laboratory you are asked to make a 0.150 m manganese(II) nitrate solution using 13.9 grams of manganese(II) nitrate.
How much water should you add? =
grams
(2a) An aqueous solution is 16.0% by mass potassium bromide, KBr, and has a density of 1.12 g/mL.
The molality of potassium bromide in the solution is= m.
(b) An aqueous solution of iron(II) sulfate has a concentration of 0.221 molal.
The percent by mass of iron(II) sulfate in the solution is = %.
(c) An aqueous solution is 13.0% by mass potassium bromide, KBr, and has a density of 1.10 g/mL.
The molality of potassium bromide in the solution is = m
Explanation / Answer
1)
a)
Molar mass of NaBr,
MM = 1*MM(Na) + 1*MM(Br)
= 1*22.99 + 1*79.9
= 102.89 g/mol
mass(NaBr)= 26.7 g
use:
number of mol of NaBr,
n = mass of NaBr/molar mass of NaBr
=(26.7 g)/(1.029*10^2 g/mol)
= 0.2595 mol
m(solvent)= 427 g
= 0.427 kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.2595 mol)/(0.427 Kg)
= 0.6077 molal
Answer: 0.608 molal
b)
m(solvent)= 475 g
= 0.475 kg
use:
number of mol,
n = Molality * mass of solvent in Kg
= (0.704 mol/Kg)*(0.475 Kg)
= 0.3344 mol
Molar mass of MnI2,
MM = 1*MM(Mn) + 2*MM(I)
= 1*54.94 + 2*126.9
= 308.74 g/mol
use:
mass of MnI2,
m = number of mol * molar mass
= 0.3344 mol * 3.087*10^2 g/mol
= 1.032*10^2 g
Answer: 103 g
c)
Molar mass of Mn(NO3)2,
MM = 1*MM(Mn) + 2*MM(N) + 6*MM(O)
= 1*54.94 + 2*14.01 + 6*16.0
= 178.96 g/mol
mass(Mn(NO3)2)= 13.9 g
use:
number of mol of Mn(NO3)2,
n = mass of Mn(NO3)2/molar mass of Mn(NO3)2
=(13.9 g)/(1.79*10^2 g/mol)
= 7.767*10^-2 mol
use:
Molality,
m = number of mol / mass of solvent in Kg
0.150 = (7.767*10^-2 mol) / mass of solvent in Kg
mass of solvent = 0.5178 Kg
mass of solvent = 517.8 g
Answer: 518 g
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