Need help in molality! full points for best answer!

[From: ] [author: ] [Date: 12-04-03] [Hit: ]

27/1.45 = .......

A solution is prepared from 16 grams of acetic acid (CH[3]CO[2]H) and 1450 grams of water. What is the solution's molality?

molality = moles of solute/kilograms of solute

16g CH3CO2H * (mol/60g) = .27 moles

1450 grams H2O * (1 kg/1000g) = 1.45kg

.27/1.45 = .186 m

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