We have that the molecular weight (3sf) of the compound (g/mol)
[tex]m=44.15g/mol[/tex]
From the question we are told
A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).
Generally the equation for the Rouault's law is mathematically given as
P=P_0 N
[tex]22.67=23.8*\frac{\frac{12.5}{18}}{\frac{125}{18}+\frac{15}{m}}\\\\\6.95+\frac{15}{m}=7.29\\\\\frac{15}{m}=7.29-6.95\\\\m=\frac{15}{0.34}\\\\m=44.11g/mol[/tex]
Therefore
The molecular weight (3sf) of the compound (g/mol)
[tex]m=44.15g/mol[/tex]
For more information on this visit
https://brainly.com/question/17756498
