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A solution of LiCl in water has XLiCl.0800. What is the Molality 4.01 m LiCl 4.44 m LiCl 4.83 m LiCl 8.70 m LiCl?
Wiki User
∙ 14y ago
C = 4.83 m LiCl
mole ratio = X
moles = n
A = solute
B = solvent
XA = nA___
nA + nB
solution of LiCl in water
solute = LiCl
solvent = water = H2O
molality = moles of solute = moles of solute
kg of solvent 1.0 kg of solvent
Assume 1.000 kg H2O
1.000 kg H2O * 1000 g * 1 mol H2O = 55.49 mol H2O
1 kg 18.02 g H2O
XA = nA___
nA + nB
Solve for nA
XA( nA + nB ) = nA
XAnA = XBnB = nA
XAnB = nA - XAnA
XAnB = nA( 1 - XA)
XAnB___ = nA
( 1 - XA)
Plug in known amounts
nA = 0.08 * 55.49 mol H2O = 4.83 mol LiCl
1 - 0.08
If molality = moles of solute
1.0 kg of solvent
THEN molality of LiCl = 4.83 mol LiCl = 4.83 m
1.000 kg H2O
Wiki User
∙ 14y ago
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