Answer :
Answer: The freezing point and boiling point of the solution are [tex]-6.6^0C[/tex] and [tex]101.8^0C[/tex] respectively.
Explanation:
Depression in freezing point:
[tex]T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_f[/tex] = freezing point of solution = ?
[tex]T^o_f[/tex] = freezing point of water = [tex]0^0C[/tex]
[tex]k_f[/tex] = freezing point constant of water = [tex]1.86^0C/m[/tex]
i = vant hoff factor = 1 ( for non electrolytes)
m = molality
[tex]w_2[/tex] = mass of solute (ethylene glycol) = 21.4 g
[tex]w_1[/tex]= mass of solvent (water) = [tex]density\times volume=1.00g/ml\times 97.6ml=97.6g[/tex]
[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62g/mol
Now put all the given values in the above formula, we get:
[tex](0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}[/tex]
[tex]T_f=-6.6^0C[/tex]
Therefore,the freezing point of the solution is [tex]-6.6^0C[/tex]
Elevation in boiling point :
[tex]T_b-T^b^0=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]100^0C[/tex]
[tex]k_b[/tex] = boiling point constant of water = [tex]0.52^0C/m[/tex]
i = vant hoff factor = 1 ( for non electrolytes)
m = molality
[tex]w_2[/tex] = mass of solute (ethylene glycol) = 21.4 g
[tex]w_1[/tex]= mass of solvent (water) = [tex]density\times volume=1.00g/ml\times 97.6ml=97.6g[/tex]
[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62g/mol
Now put all the given values in the above formula, we get:
[tex](T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}[/tex]
[tex]T_b=101.8^0C[/tex]
Thus the boiling point of the solution is [tex]101.8^0C[/tex]