Answer: [tex]2.23^0C/m[/tex]
Explanation:
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_{solution}-T_{solvent}=k_b\times \frac{\text{ Moles of solute}\times 1000}{\text{ Mass of solvent in g}}[/tex]
where,
[tex]T_b[/tex] = change in boiling point
[tex]k_b[/tex] = boiling point constant
m = molality
moles of solute 9urea)=[tex]\frac{given mass}{Molar mass}=\frac{58.66}{60.06g/mol}=0.98moles[/tex]
Putting in the values we get:
[tex](123-120.7)^0C=k_b\times \frac{0.98}\times 1000}{950.0}[/tex]
[tex]2.3^0C=k_b\times \frac{0.98}\times 1000}{950.0}[/tex]
[tex]k_b=2.23^0C/m[/tex]
Thus molal boiling point elevation constant is [tex]2.23^0C/m[/tex]
