[bpeck8]

This is the question:

What mass of napthalene, C10H8, must be dissolved in 200.g of benzene to produce a solution with a freezing point 1.70 C below that of pure benzene? Kf for benzene is 5.12 C Kg/mol.

My work:

Tf=Kfm, Tf=m
            Kf   

(200g benzene)(0.001Kg) = .2Kg benzene
                        1g

1.70 C          = moles of C10H8
5.12 C Kg/mol    .2Kg benzene

(.2Kg)(.332Kg/mol)= moles of C10H8

.0664mol=moles of C10H8

(.0664mol)(104g  ) = 6.91 g H10H8
                1 mol

[Schrödinger]

This is the question:

What mass of napthalene, C10H8, must be dissolved in 200.g of benzene to produce a solution with a freezing point 1.70 C below that of pure benzene? Kf for benzene is 5.12 C Kg/mol.

My work:

Tf=Kfm, Tf=m
            Kf   

(200g benzene)(0.001Kg) = .2Kg benzene
                        1g

1.70 C          = moles of C10H8
5.12 C Kg/mol    .2Kg benzene

(.2Kg)(.332Kg/mol)= moles of C10H8

.0664mol=moles of C10H8

(.0664mol)(104g  ) = 6.91 g H10H8
                1 mol

Molecular mass of naphthalene = 104? Please recheck that.

[bpeck8]

hehe yeah, made that mistake. 128

answer turns out to be 8.45

thanks for the help