Question

What is the boiling point of a solution that is produced by dissolving 20g NaOH in 2L H2O if the Kb of H2O is 0.52 oC/m?

a. 100.0oC
b. 100.3 oC
c. 100.4 oC
d. 101.0 oC

Answer 1

$\Delta T=K_b *m_l *i$ , $\Delta T$ - temperature change (T₂-T₁), K_b - the molal boiling point elevation constant, m_l - molality or molal concentration of the solute in the solution, i - isotonic coefficient that takes into account dissociation; $i = 1+\alpha (\nu -1)$, $\alpha$ - degree of dissociation, $\nu$ - the number of ions formed during dissociation. $n=m/M$ , n - chemical amount of substance; m - mass of substance; M - molar mass of substance. M(NaOH)=23g/mol+16g/mol+1g/mol=40g/mol.n(NaOH)=20g/40g/mol=0,5mol$m_l =n/m(H_2O)$ , n - chemical amount of substance; m(H₂O) - solvent mass(kg), in our case - H₂O. $m=V*\rho$ , m - mass, V - volume, $\rho$ - density (for H₂O is 997kg/m³).2L=2dm³=0,002m³ m(H2O)=0,002m3*997kg/m3=1,994kgml=0,5mol/1,994kg=0,250mol/kg$i$ = 1+1(2-1)=2 ($NaOH\to Na^+ +OH^-$ , $\nu$ =2, $\alpha$ =1, because NaOH is a strong electrolyte)

$\Delta T = 0,52oC/m*0,250mol/kg*2=0,26$oC

T₁=100oC (water boiling point), $T_2 =T_1 +\Delta T$ , T₂=100,26 round the number to 100,3.

Answer: b.