Asked by deepak
lowest+freezing+point+of+Kbr+AlNO33+CH3COONa+NaNO2+MgCl2+of+same+molality
Answers
Answered by
DrBob222
You would do well to make your post easier to read and understand.
All you need is
delta T = i*Kf*m
i = 2 for KBr.
i = 4 for Al(NO3)3
i = 2 for CH3COONa
i = 2 for NaNO2
i = 3 for MgCl2
If m is constant and Kf is constant, then delta T = some constant x i so the one with the most particles will have the lowest freezing point.
All you need is
delta T = i*Kf*m
i = 2 for KBr.
i = 4 for Al(NO3)3
i = 2 for CH3COONa
i = 2 for NaNO2
i = 3 for MgCl2
If m is constant and Kf is constant, then delta T = some constant x i so the one with the most particles will have the lowest freezing point.
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