Question

1. what is the molar mass of X(i = 1) if 50.00 water containing 2.50 g of X has a freezing point of -11.5 degree Celcius

2.Using information from Question 1, recalculate the molar mass of X if i = 2

Answer #1

**normal boiling point of water solution = 0
C**

**observed boiling point of water solution = -11.5
C**

**depression in freezing poin dTf = 0 - ( -11.5
)**

**dTf = 11.5 C**

**we know that**

**depression in freezing point (dTf) = i x Kf x
m**

**for solvent water Kf = 1.86**

**11.5 = 1 x 1.86 x m**

**m = 6.18**

**now**

**molality = moles of solute / mass of solvent
(kg)**

**6.18 = moles of X / 0.05**

**moles of X = 0.309**

**now**

**molar mass = mass /moles**

**molar mass of X = 2.5 / 0.309**

**molar mas of X = 8.08 g**

**2)**

**normal boiling point of water solution = 0 C**

**observed boiling point of water solution = -11.5
C**

**depression in freezing poin dTf = 0 - ( -11.5
)**

**dTf = 11.5 C**

**we know that**

**depression in freezing point (dTf) = i x Kf x
m**

**for solvent water Kf = 1.86**

**11.5 = 2 x 1.86 x m**

**m = 3.09**

**now**

**molality = moles of solute / mass of solvent
(kg)**

**3.09 = moles of X / 0.05**

**moles of X = 0.1545**

**now**

**molar mass = mass /moles**

**molar mass of X = 2.5 / 0.1545**

**molar mas of X = 16.17 g**

ASAP! PLZ I need all answers for pre lab question in
Molecular Weight Determination by Freezing Point
Depression
Prelab questions:
(PL1) What is the molar mass of X (i=1) if 50.00g of water
containing 2.50g of X has a freezing point of -11.5C?
(PL2) Using information from PL5, recalculate the molar mass of
X if i=2.
(PL3) Compare your answers from (PL5) and (PL6). Briefly
explain.
(PL4) How is melting point different than freezing point?
(PL5) List the following in...

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m=molality = moles solute/kg solvent Molar mass= grams of
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