Respuesta :
Answer : The the concentration of [tex]CaCl_2[/tex] in mass percent is, 41.18 %
Solution : Given,
Molar mass of water = 18 g/mole
Molar mass of [tex]CaCl_2[/tex] = 110.98 g/mole
First we have to calculate the mole fraction of solute.
According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.
[tex]\fac{p^o-p_s}{p^o}=X_B[/tex]
where,
[tex]p^o[/tex] = vapor pressure of the pure component (water) = 92.6 mmHg
[tex]p_s[/tex] = vapor pressure of the solution = 83.1 mmHg
[tex]X_B[/tex] = mole fraction of solute, [tex](CaCl_2)[/tex]
Now put all the given values in this formula, we get the mole fraction of solute.
[tex]\fac{92.6-83.1}{92.6}=X_B[/tex]
[tex]X_B=0.102[/tex]
Now we have to calculate the mole fraction of solvent (water).
As we know that,
[tex]X_A+X_B=1\\\\X_A=1-X_B\\\\X_A=1-0.102\\\\X_A=0.898[/tex]
The number of moles of solute and solvent will be, 0.102 and 0.898 moles respectively.
Now we have to calculate the mass of solute, [tex](CaCl_2)[/tex] and solvent, [tex](H_2O)[/tex].
[tex]\text{Mass of }CaCl_2=\text{Moles of }CaCl_2\times \text{Molar mass of }CaCl_2[/tex]
[tex]\text{Mass of }CaCl_2=(0.102mole)\times (110.98g/mole)=11.32g[/tex]
[tex]\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O[/tex]
[tex]\text{Mass of }H_2O=(0.898mole)\times (18g/mole)=16.164g[/tex]
Mass of solution = Mass of solute + Mass of solvent
Mass of solution = 11.32 + 16.164 = 27.484 g
Now we have to calculate the mass percent of [tex]CaCl_2[/tex]
[tex]Mass\%=\frac{\text{Mass of}CaCl_2}{\text{Mass of solution}}\times 100=\frac{11.32g}{27.484g}\times 100=41.18\%[/tex]
Therefore, the the concentration of [tex]CaCl_2[/tex] in mass percent is, 41.18 %
