68.0 g of a solute were dissolved in 393 mL of benzene. The solution froze at -0.50°C. The normal freezing point of benzene is 5.5°C and the molar freezing point constant is 5.12°C kg/mol. The density of benzene is 0.879 g/mL. Calculate the molar mass of the solute.
i am so clueless. i am in school to be a chiropractor and the are making me take every chemistry course ever created, basically. i know the molar mass equation is N=M/m, but this problem gives me all kinds of temperature and stuff that i don't know what to do with...please help.
No, i am not asking you to do my homework. there are about 12 problems just like this on my worksheet and i just need someone to show me how to do one to get started so i can hopefully teach myself the other ones.
i am so clueless. i am in school to be a chiropractor and the are making me take every chemistry course ever created, basically. i know the molar mass equation is N=M/m, but this problem gives me all kinds of temperature and stuff that i don't know what to do with...please help.
No, i am not asking you to do my homework. there are about 12 problems just like this on my worksheet and i just need someone to show me how to do one to get started so i can hopefully teach myself the other ones.
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mass solute = 68.0 g
kg solvent = 393 mL(0.879 g/mL)(1 kg)/(1000 g) = 0.3454 kg
molality = m = moles solute/kg solvent = (mass solute)/(MW)/(kg solvent)
You still have 2 variables and one equation so you need to find molality.
ΔT = -Kf*m), where [ΔT = T(final) - T(initial) = -0.50°C - (5.50°C) = -6.00°C]
m = ΔT/-Kf = (-6.00°C)/(-5.12°C kg/mol) = (-6.00°C)/(-5.12°C/molal) = 1.17 m
Now, back to the equation that has MW in it.
molality = m = (mass solute)/(MW)/(kg solvent)
1.17 m = [(68.0 g)/MW]/(0.3454 kg) --> now solve for MW
MW = (mass solute)/(molality)(kg solvent) = (68.0 g)/(1.17 m)(0.3454 kg)
MW = 168 g/mol
Hope this is helpful to you. JIL HIR
kg solvent = 393 mL(0.879 g/mL)(1 kg)/(1000 g) = 0.3454 kg
molality = m = moles solute/kg solvent = (mass solute)/(MW)/(kg solvent)
You still have 2 variables and one equation so you need to find molality.
ΔT = -Kf*m), where [ΔT = T(final) - T(initial) = -0.50°C - (5.50°C) = -6.00°C]
m = ΔT/-Kf = (-6.00°C)/(-5.12°C kg/mol) = (-6.00°C)/(-5.12°C/molal) = 1.17 m
Now, back to the equation that has MW in it.
molality = m = (mass solute)/(MW)/(kg solvent)
1.17 m = [(68.0 g)/MW]/(0.3454 kg) --> now solve for MW
MW = (mass solute)/(molality)(kg solvent) = (68.0 g)/(1.17 m)(0.3454 kg)
MW = 168 g/mol
Hope this is helpful to you. JIL HIR