Respuesta :
The vapor pressure of a solution will decrease and the freezing point will decrease with an increasing in the amount of dissolved solute. So the answer is third option.
Answer : The correct option is, (3) decrease, decrease, increase
Explanation :
As the relative lowering of vapor pressure and the lowering of freeing point is depends upon the amount of dissolved solute. That the relative lowering of vapor pressure and the lowering of freeing point is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}[/tex]
where,
[tex]p^o[/tex] = vapor pressure of pure solvent
[tex]p_s[/tex] = vapor pressure of solution
[tex]w_2[/tex] = mass of solute
[tex]w_1[/tex] = mass of solvent
[tex]M_1[/tex] = molar mass of solvent
[tex]M_2[/tex] = molar mass of pure solvent
As the amount of solute dissolved increases, the relative lowering of vapor pressure also increases but the vapor pressure of a solution will decreases because the vapor pressure of the solution is lower than that of the pure solvent.
The formula for relative lowering of freezing pressure will be,
[tex]\Delta T_f= k_f\times m\\\\T^o_f-T_f= k_f\times \frac{w_2\times 1000}{M\times w_1}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point or freezing point depression
[tex]T^o_f[/tex] = freezing point of pure solvent
[tex]T_f[/tex] = freezing point of solution
[tex]k_f[/tex] = freezing point constant
m = molality
As the amount of solute dissolved increases, the relative lowering of freezing point also increases but the freezing of a solution will decreases because the freezing point of the solution is lower than that of the pure solvent.
Hence, the vapor pressure of a solution will decreases and the freezing point will decreases with an increase in the amount of dissolved solute.
