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I know that molality is 0.41 but how do I get freezing and boiling point?

I used this equation but I kept getting wrong answer. Can you show mw how?

What numbers are you using for your values?

Kb= 2.53 C/m
Kr = 5.1 C/m

Your first step is to convert your grams of naphthalene and benzene into moles

So 2.20g *(1 mol/128.17g/mol)=?
42.2 * (1mol/78.11g/mol)=?

I am still confused.

0.017 mol o naphthalene
0.0422 kg of benzene

Can you show me how? I am still confused.

how did you do? what is your pure solvent and what is the molality?

sorry I see the molality is correct 10.74 m

what are your values of k for the water?

now multiply the values of k times the molality

Freezing point 0-(1.86x10.74)= ???
Boiling point 100+ (0.512 x 10.74) = ?????