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 2 of 3.
Hareesh Somineni said:
1 decade 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)
Satish said:
1 decade ago
Molality = Number of grm moles of solute per kilogram of solvent.
= 0.3*1000/600=0.5m
= 0.3*1000/600=0.5m
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)
Rajesh said:
1 decade ago
Molality = Number of gram moles of solute per kilogram of solvent.
Poonam Rathod said:
1 decade ago
Molality = (solute/solvent) *100.
= (0.30/600) *1000.
= 0.50
= (0.30/600) *1000.
= 0.50
(1)
Ram said:
1 decade ago
Molality= (Wt.of Sloute*1000)/(Mol.wt * Wt of the slovent)
(1)
Anurag said:
1 decade ago
Gram mole of solute per kilogram of solvent = Molality.
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)
Kena Muluna Ayansal said:
2 years ago
Thanks, everyone for explaining the answer.
(1)
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