Answer:
139.94 grams of solute is present in kilograms of water
Explanation:
Temperature of the ice water = T = 1.0°C
Temperature of the mixture = [tex]T_f[/tex] =-3.0°C
Depression in freezing point [tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f[/tex] =1.0°C-( -3.0°C)=4°C
[tex]\Delta T_f=k_f\times m=k_f\times \frac{\text{Moles of solute}}{\text{mass of solvent (kg)}}[/tex]
m = molality of the solution
[tex]k_f[/tex] = Molal depression constant
[tex]4^oC=1.86^oC kg/mol\times \frac{\text{Moles of solute}}{0.0793 kg}[/tex]
Mole of solute = 0.1705 moles
Mass of solute = 11.1 g
Moles of solute = [tex]0.1705 mol=\frac{11.1 g}{M}[/tex]
Molar mass of the solute= M
M = 65.088 g/mol
Molality of the solution = m = [tex]\frac{0.1705 mol}{0.0793 kg}[/tex]
[tex]m\times M=\frac{0.1705 mol}{0.0793 kg}\times 65.088 g/mol=139.94 [/tex]g of solute /kg of water
139.94 grams of solute is present in kilograms of water