.20 \mathrm{~mm. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions,Colligative Properties | Abnormal Colligative Properties | Rlvp,Raoult's Law, Class 11 Chemistry,Class 12 Chemistry,JEE Main Ch.">
.20 \mathrm{~mm. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions,Colligative Properties | Abnormal Colligative Properties | Rlvp,Raoult's Law, Class 11 Chemistry,Class 12 Chemistry,JEE Main Ch.">
.20 \mathrm{~mm. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions,Colligative Properties | Abnormal Colligative Properties | Rlvp,Raoult's Law, Class 11 Chemistry,Class 12 Chemistry,JEE Main Ch.">
What weight of glucose must be dissolved in $100 \mathrm{~g}$ of water to lower the vapour pressure by $0.20 \mathrm{~mm} ~\mathrm{Hg}$ ?
(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is $54.2 \mathrm{~mm} ~\mathrm{Hg}$ at room temperature. Molar mass of glucose is $180 \mathrm{~g} \mathrm{~mol}^{-1}$
3.69 g
2.59 g
3.59 g
4.69 g
Solution
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Step-by-Step Solution
Step 1: State Raoult's Law for lowering of vapor pressure
According to Raoult’s Law, the lowering of vapor pressure ($\Delta P$) of a solvent upon adding a non-volatile solute is directly proportional to the mole fraction of the solute ($x_{\text{solute}}$):
$ \Delta P = x_{\text{solute}} \cdot P_0
$
where $P_0$ is the vapor pressure of the pure solvent.
Step 2: Express the mole fraction of the solute
From the data:
Lowering of vapor pressure, $\Delta P = 0.20 \,\mathrm{mm\,Hg}$
Vapor pressure of pure water, $P_0 = 54.2 \,\mathrm{mm\,Hg}$
The mole fraction of the solute (glucose) is given by:
$ x_{\text{solute}}
= \frac{\Delta P}{P_0}
= \frac{0.20}{54.2}
\approx 0.003689
$
Step 3: Relate mole fraction to number of moles
For a dilute solution, the total number of moles in the solution is approximately the number of moles of the solvent (water), since the solute amount is small. Therefore:
$ x_{\text{solute}}
= \frac{\text{moles of solute}}{\text{moles of water}}
$
Rearranging for the moles of solute gives:
$ \text{moles of solute}
= x_{\text{solute}} \times \text{moles of water}.
$
Step 4: Calculate the moles of water
The mass of water is $100\,\mathrm{g}$, and the molar mass of water is $18\,\mathrm{g\,mol}^{-1}$. Hence:
$ \text{moles of water}
= \frac{100\,\mathrm{g}}{18\,\mathrm{g\,mol}^{-1}}
\approx 5.56\,\mathrm{mol}.
$
Step 5: Determine the moles of glucose
Substitute the values into the expression derived for the moles of solute:
$ \text{moles of glucose}
= 0.003689 \times 5.56\,\mathrm{mol}
\approx 0.0205\,\mathrm{mol}.
$
Step 6: Convert moles of glucose to mass
The molar mass of glucose is $180\,\mathrm{g\,mol}^{-1}$. Therefore, the required mass of glucose is:
$ \text{mass of glucose}
= 0.0205\,\mathrm{mol} \times 180\,\mathrm{g\,mol}^{-1}
\approx 3.69\,\mathrm{g}.
$
Step 7: Final result
Thus, the mass of glucose required to be dissolved in $100\,\mathrm{g}$ of water to lower its vapor pressure by $0.20\,\mathrm{mm\,Hg}$ is
3.69 g.