Chemical Engineering - Mass Transfer - Discussion
Discussion Forum : Mass Transfer - Section 1 (Q.No. 2)
2.
In a solution containing 0.30 Kg mole of solute and 600 kg of solvent, the molality is
Discussion:
21 comments Page 1 of 3.
Om pal said:
1 year ago
Molality = gram moles of Salute/mass of solute in kg.
Then, 0.30 * 1000/600 = 300.00/600 = 1/2 = 0.50 answer.
Then, 0.30 * 1000/600 = 300.00/600 = 1/2 = 0.50 answer.
(3)
Ramsajan paswan said:
2 years ago
Molality = gram moles of Salute/mass of solute in kg.
Then 0.30 * 1000/600 =0.50.
Then 0.30 * 1000/600 =0.50.
(16)
Kena Muluna Ayansal said:
2 years ago
Thanks, everyone for explaining the answer.
(1)
Meghan Desai said:
3 years ago
Molality = moles of solute (in moles)/kg of solvent.
(2)
Varun Kr Sr said:
3 years ago
Molality = g-mol of solute÷Mass of solvent in kg.
(1)
PATEL ANAND said:
5 years ago
Morality is defined as,
Molality=(solute (gram)/mass of solvent in kg)
=> volume density of solution is kg/10000.
So, 600/1000= 0.600.
M=0.3 kg/0.600liter=0.30*0.600=0.50 kmol/litre.
Molality=(solute (gram)/mass of solvent in kg)
=> volume density of solution is kg/10000.
So, 600/1000= 0.600.
M=0.3 kg/0.600liter=0.30*0.600=0.50 kmol/litre.
(1)
Amit Kaushik said:
7 years ago
Molality = (gmol of solute/Mass of Solvent in Kg).
So, Molality = (.3 * 1000)/600 = 0.5.
So, Molality = (.3 * 1000)/600 = 0.5.
(1)
Ahmed Gamal Fayed said:
7 years ago
Density of water = 1000 kg/m3 {g/liter}.
Volume of solution = kg/ density.
= 600/ 1000 = 0.6 liter.
M = 0.3 kg/0.6 liter = 0.50 kg mole/ liter.
Volume of solution = kg/ density.
= 600/ 1000 = 0.6 liter.
M = 0.3 kg/0.6 liter = 0.50 kg mole/ liter.
(2)
Arshad Ali said:
7 years ago
Molality=(mol of solute/Kg of solvent).
So, 0.30 kg mol=0.30*1000 mol= 300 mol,
Hence, Molality = 300mol/600Kg=0.5 mol/Kg.
So, 0.30 kg mol=0.30*1000 mol= 300 mol,
Hence, Molality = 300mol/600Kg=0.5 mol/Kg.
(7)
Raj said:
8 years ago
Molality = (kg solute/kg solvent)*100.
= (.3/600)*1000.
= 0.5.
= (.3/600)*1000.
= 0.5.
(2)
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