Freezing Point Depression Problem

[ebon sequoia]

Would really appreciate some help for this problem. Im in a big gen chem 2 class and alot of us are all getting different answers and we dont know why. Some of us got 0.034 and some 0.04, and someone else said the answer was 0.0. from my understanding the solution goes like this:

• moles of naphthalene: 6.0 g/128.17 g/mol = 0.0468 mol
• mass of benzene: 50.0 mL * 0.877 g/mL = 43.85 g
• molality: 0.0468 mol/0.04385 kg = 1.067 mol/kg
• freezing point: 5.12 C/m * 1.067 mol/kg = 5.46 C
• New freezing point: 5.5 C - 5.46 = 0.04 C

[turbid root]

mm maybe you can check #old-network for a chemistry server to make sure

From what I understand, you want 5.5 - K*molality

,calc 5.5 - 5.12* ((6/128.17)/(50*0.877/1000))

[glossy fulcrum]

Result:

0.034055352831441