Calculate the boiling point of a solution prepared by dissolving 70.0 g of naphthalene, C10H8 (a nonvolatile nonelectrolyte), in 220.0 g of benzene, C6H6. The Kb for benzene = 2.53oC/m. The boiling point of pure benzene is 80.1oC.

Ans: 86.4 degrees Celsius
I did this so far
1)70G C10H8(1mol c10H8/128gC10H8)= .546mol C10H8
2)(.546mol c10h8)(.220kg benzene)=2.482m
3) (2.482)(2.53 C/m)=6.289
I'm not sure if I am starting this off right, can anyone help me get the correct answer? Please ans ty!

Question

Calculate the boiling point of a solution prepared by dissolving 70.0 g of naphthalene, C10H8 (a nonvolatile nonelectrolyte), in 220.0 g of benzene, C6H6. The Kb for benzene = 2.53oC/m. The boiling point of pure benzene is 80.1oC.

Ans: 86.4 degrees Celsius
I did this so far
1)70G C10H8(1mol c10H8/128gC10H8)= .546mol C10H8
2)(.546mol c10h8)(.220kg benzene)=2.482m
3) (2.482)(2.53 C/m)=6.289
I'm not sure if I am starting this off right, can anyone help me get the correct answer? Please ans ty!

chmvijay · Answer

ΔT =(Kb×1000×W2)/M2×W1
        = 2.53 *1000*70 /128.17 * 220
       =177100 /28197.4
        =6.28
but 
ΔT=T-To
6.28 =T - 80.1
T=6.28+80.1 =86.38 °C

chmvijay · Answer

where W2 - gram of non volatile solute naphthalene
W1 - gram of solvent benzene,
M2 molar mass of naphthalene ,