Calculate the mole fraction, molarity and molality of NH3 if it is in
solution composed of 30.6 g NH3 in 81.3 g of H2O. The density of the
solution is 0.982 g/mL and the density of water is 1.00 g/mL
Answers
Answer:
-Molar mass of NH₂ = 179
•Moler mass of H₂O = 18g
No of moles in 30.6g NH3:
= 30.6/17
= 1,8 moles
No of moles in 81.3g H₂O
= 81.3/18
= 4.5 moles
Molality = Moles of solute/Mass of solvent(kg)
= 18/[813/1000]
= 1.8/0.0813
= 22.1 m
Mole fraction:
= Moles of solute/Total moles in solution
= 18/1.8+4.5]
18/6.3
= 0.285
We know the relationship that:
>>>m= 1000M/1000d - MM
Where:
mis molality
Mis molarity
-d is density of solvent (in g/mL)
•M, is molar mass of solute
⇒ 22:1=1000M/1000(0.982)-M(17)
221-1000M/982-17M
1000M-22.1982-17M)
1000M-21702 2-375.7
>>217022=1375.7M => M=21702.2/1375.7
⇒>M-15.8
Answer:
MARK ME AS THE BRAINIEST ANSWER ❤️❤️ (✷‿✷)
First of all let's calculate the molar weight of
NH3 =14+ 3= 17g
H2O =2 + 16= 18g
Now let's calculate the number of moles of
NH3 in 30.6g by dividing it by molar weight = 30.6/17 = 1.8 moles
Similarly, the number of moles of H20 in 81.3g = 81.3/18 = 4.5
Mole fraction = number of moles of solute/number of moles of solvent = 1.8/4.5 = 0.4
Molality = number of moles of solute /weight of solvent in kg = 1.8/81.3 X 1000 = 22 moles per gm of water
Molarity = number of moles of solute / volume of solvent in L = 1.8/81.3 × 1000 = 22 M per mL of water
Now except for the units, the molarity and molality is the same in this case, this is because, density of water is 1g per mL at room temperature.