Respuesta :
Answer: The molar mass of the solute is 300 g/mol
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 : 10.0 g of non volatile solute is present in 78.11 g of solvent benzene.
moles of solute = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{10.0g}{Mg/mol}[/tex]
moles of solvent (benzene) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{78.11g}{78.11g/mol}=1mole[/tex]
Total moles = moles of solute (glycerol) + moles of solvent (water) = [tex]\frac{10.0}{M}+1[/tex]
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{10.0}{M}}{\frac{10.0}{M}+1}[/tex]
[tex]\frac{0.930-0.900}{0.930}=1\times \frac{\frac{10.0}{M}}{\frac{10.0}{M}+1}[/tex]
[tex]M=300g/mol[/tex]
Thus the molar mass of the solute is 300 g/mol
The molar mass of the solute is 300 grams per mole.
Calculation of the molar mass:
Based on the given information,
The vapor pressure of the pure solvent (Po) is 0.930 atm.
The vapor pressure of the solution (Ps) is 0.900 atm.
The mass of benzene given is 78.11 grams.
The mass of the nonvolatile solute is 10 grams.
Let the molar mass of the solute be x. The moles of solute can be determined by using the formula,
n = Weight/Molecular mass
Now the moles of solute will be,
[tex]n = \frac{10 g}{x}[/tex]
Now the molecular mass of benzene is 78 g/mol, and the mass of benzene given is 78.11 grams. The number of moles of benzene will be,
N = Weight of benzene/Molecular mass of benzene
Putting the values we get,
[tex]N = \frac{78.11 g}{78 g/mol} \\N = 1.00 moles[/tex]
Now based on the Raoult's law,
[tex]\frac{Po - Ps}{Po} = \frac{n}{n+N}[/tex]
Now putting the values we get,
[tex]\frac{0.930 - 0.900}{0.930} = \frac{10/x}{10/x +1.00} \\0.032258 = \frac{10}{10+1.00 x} \\x = 300 g/mol[/tex]
Thus, the molar mass of the solute is 300 g/mol.
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