Respuesta :
Answer : The boiling point of the resulting solution is, [tex]100.6^oC[/tex]
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=i\times k_b\times m[/tex]
or,
[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b[/tex] = boiling point of solution = ?
[tex]T^o_b[/tex] = boiling point of water = [tex]100^oC[/tex]
[tex]k_b[/tex] = boiling point constant = [tex]0.52^oC/m[/tex]
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
[tex]w_2[/tex] = mass of solute (sucrose) = 5.0 g
[tex]w_1[/tex] = mass of solvent (water) = 10.0 g
[tex]M_2[/tex] = molar mass of solute (sucrose) = 342.3 g/mol
Now put all the given values in the above formula, we get:
[tex](T_b-100)^oC=1\times (0.52^oC/m)\times \frac{(5.0g)\times 1000}{342.3\times (10.0g)}[/tex]
[tex]T_b=100.6^oC[/tex]
Therefore, the boiling point of the resulting solution is, [tex]100.6^oC[/tex]
