Answer:
Mass of the salt: 105.6g of KCl.
Mass water: 958.9g of water.
Molality: 1.478m.
Explanation:
<em>Mass of the salt:</em>
In 1L, there are 1.417 moles. In grams:
1.417 moles KCl * (74.54g / mol) = 105.6g of KCl
<em>Mass of the water:</em>
We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:
1L * (1064.5g / L) = 1064.5g - Mass solution.
Mass water = 1064.5g - 105.6g = 958.9g of water
<em>Molality:</em>
Moles KCl = 1.417 moles KCl.
kg Water = 958.9g = 0.9589kg.
Molality = 1.417mol / 0.9589kg = 1.478m
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Answer:
You can find in the given attachemnet
<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
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