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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

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Q: 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?
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