Find the boiling-point elevation of the water solution of NaCl, if in 100 g of water there is 39 g of NaCl. Find the boiling-point elevation and the boiling-point of the solution.
m(NaCl) = 39 g = 0.039 kg
m (H2O) = 100 g = 0.1 kg
m(NaCl(aq)) = 139 g = 0.139 kg
Kb(H2O) = 0.512 Kkg/mol
Tv=?
?T=i * Kb * b
i = 2
Kb = 0.512 Kkg/mol
b = n(NaCl)/m(H2O) = 0.67 mol / 0.1 kg = 6.7 mol/kg
n(NaCl)= m(NaCl)/M(NaCl) = 39 g / 58.44 g/mol = 0.67 mol
?Tv=2 * 0.512 Kkg/mol * 6.7 mol/kg
?Tv = 6.86 K
= -266°C ?
Is that correct? I'm not sure.
m(NaCl) = 39 g = 0.039 kg
m (H2O) = 100 g = 0.1 kg
m(NaCl(aq)) = 139 g = 0.139 kg
Kb(H2O) = 0.512 Kkg/mol
Tv=?
?T=i * Kb * b
i = 2
Kb = 0.512 Kkg/mol
b = n(NaCl)/m(H2O) = 0.67 mol / 0.1 kg = 6.7 mol/kg
n(NaCl)= m(NaCl)/M(NaCl) = 39 g / 58.44 g/mol = 0.67 mol
?Tv=2 * 0.512 Kkg/mol * 6.7 mol/kg
?Tv = 6.86 K
= -266°C ?
Is that correct? I'm not sure.