Respuesta :
Answer: 158 grams
Explanation:
As the relative lowering of vapor pressure 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}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor = 1 (for non electrolytes)
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]
Given : x g of ethylene glycol is present in 183 g of water
moles of solute (ethylene glycol) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{x}{62.07}moles[/tex]
moles of solvent (water) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{183g}{18.02g/mol}=10.2moles[/tex]
Total moles = moles of solute (ethylene glycol) + moles of solvent (water) = [tex]\frac{x}{62.07}[/tex] + 10.2
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{x}{62.07}}{\frac{x}{62.07}+10.2}[/tex]
[tex]\frac{1.00-0.800}{1.00}=1\times \frac{\frac{x}{62.07}}{\frac{x}{62.07}+10.2} [/tex]
[tex]0.2= \frac{\frac{x}{62.07}}{\frac{x}{62.07}+10.2}[/tex]
[tex]x=158g[/tex]
Thus the mass of ethylene glycol should be 158 g
