Respuesta :
Answer:
Solution with 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol will have lowest freezing point
Explanation:
[tex]\Delta T_f=K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point = Â
[tex]K_f[/tex] = freezing point constant Â
m = molality
As we can see that higher the molality of the solution more will depression in freezing point of the solution and hence lower will the freezing point of solution.
[tex]Molality=\frac{moles}{\text{mass of solvent in kg}}[/tex]
A. 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol.
Moles of [tex]C_3H_8O[/tex]=[tex]\frac{35.0 g}{60 g/mol}=0.5833 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m=\frac{0.5833 mol}{0.25 kg}=2.33 m[/tex]
B. 35.0 g of [tex]C4H_{10}O[/tex] in 250.0 g of ethanol
Moles of [tex]C_4H_{10}O[/tex][tex]=[tex]\frac{35.0 g}{74 g/mol}=0.4730 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m'=\frac{0.4730 mol}{0.25 kg}=1.89 m[/tex]
C. 35.0 g of [tex]C_2H_{6}O_2[/tex] in 250.0 g of ethanol
Moles of [tex]C_2H_{6}O_2[/tex]=[tex]\frac{35.0 g}{62g/mol}=0.5645 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m''=\frac{0.5645 mol}{0.25 kg}=2.26 m[/tex]
[tex]m>m'''>m''[/tex]
Solution with 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol will have lowest freezing point
