Answer: The freezing point and boiling point of the solution are
and
respectively.
Explanation:
Depression in freezing point:
![T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}](https://tex.z-dn.net/?f=T_f%5E0-T%5Ef%3Di%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7Bw_2%5Ctimes%201000%7D%7BM_2%5Ctimes%20w_1%7D)
where,
= freezing point of solution = ?
= freezing point of water = ![0^0C](https://tex.z-dn.net/?f=0%5E0C)
= freezing point constant of water = ![1.86^0C/m](https://tex.z-dn.net/?f=1.86%5E0C%2Fm)
i = vant hoff factor = 1 ( for non electrolytes)
m = molality
= mass of solute (ethylene glycol) = 21.4 g
= mass of solvent (water) = ![density\times volume=1.00g/ml\times 97.6ml=97.6g](https://tex.z-dn.net/?f=density%5Ctimes%20volume%3D1.00g%2Fml%5Ctimes%2097.6ml%3D97.6g)
= molar mass of solute (ethylene glycol) = 62g/mol
Now put all the given values in the above formula, we get:
![(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}](https://tex.z-dn.net/?f=%280-T_f%29%5E0C%3D1%5Ctimes%20%281.86%5E0C%2Fm%29%5Ctimes%20%5Cfrac%7B%2821.4g%29%5Ctimes%201000%7D%7B97.6g%5Ctimes%20%2862g%2Fmol%29%7D)
![T_f=-6.6^0C](https://tex.z-dn.net/?f=T_f%3D-6.6%5E0C)
Therefore,the freezing point of the solution is ![-6.6^0C](https://tex.z-dn.net/?f=-6.6%5E0C)
Elevation in boiling point :
![T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}](https://tex.z-dn.net/?f=T_b-T%5Eb%5E0%3Di%5Ctimes%20k_b%5Ctimes%20%5Cfrac%7Bw_2%5Ctimes%201000%7D%7BM_2%5Ctimes%20w_1%7D)
where,
= boiling point of solution = ?
= boiling point of water = ![100^0C](https://tex.z-dn.net/?f=100%5E0C)
= boiling point constant of water = ![0.52^0C/m](https://tex.z-dn.net/?f=0.52%5E0C%2Fm)
i = vant hoff factor = 1 ( for non electrolytes)
m = molality
= mass of solute (ethylene glycol) = 21.4 g
= mass of solvent (water) = ![density\times volume=1.00g/ml\times 97.6ml=97.6g](https://tex.z-dn.net/?f=density%5Ctimes%20volume%3D1.00g%2Fml%5Ctimes%2097.6ml%3D97.6g)
= molar mass of solute (ethylene glycol) = 62g/mol
Now put all the given values in the above formula, we get:
![(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}](https://tex.z-dn.net/?f=%28T_b-100%29%5E0C%3D1%5Ctimes%20%280.52%5E0C%2Fm%29%5Ctimes%20%5Cfrac%7B%2821.4g%29%5Ctimes%201000%7D%7B97.6g%5Ctimes%20%2862g%2Fmol%29%7D)
![T_b=101.8^0C](https://tex.z-dn.net/?f=T_b%3D101.8%5E0C)
Thus the boiling point of the solution is ![101.8^0C](https://tex.z-dn.net/?f=101.8%5E0C)