
The vapor pressure of pure water at $26^\circ C$ is $25.21$ torr. The vapor pressure of a solution which contains $20.0$ g glucose, ${C_6}{H_{12}}{O_6}$, in $70g$ water is:
A) $22.5$ torr
B) $23.4$ torr
C) $24.4$ torr
D) $24.5$ torr
Answer
551.1k+ views
Hint:We can solve this by understanding that the relative lowering of vapor pressure of a solution is equal to the mole fraction of the solute. Here, glucose is the solute and the amount of glucose dissolved in water is given. Calculate the mole fraction of glucose in water.
Complete step by step solution:
Given to us that $20$ grams of glucose is dissolved in $70$ grams of water.
Let us first find the number of moles of glucose present in water. Now we know that the formula for number of moles is the ratio of weight of the solute and the gram molecular weight of the solute.
Hence, the number of moles of glucose is $\dfrac{{20}}{{180}} = 0.111$
Now, let us find the number of moles of water in similar fashion.
Number of moles of water is $\dfrac{{70}}{{18}} = 3.89$
Now, the formula for mole fraction of glucose is given as the ratio of number of moles of glucose to the total number of moles.
The number of moles of glucose is $0.111$
The total number of moles is $0.111 + 3.89 = 4.001$
The formula for relative lowering of vapor pressure is given as $\dfrac{{{P^0} - P}}{{{P^0}}}$
Here, ${P^0}$ is the vapor pressure of pure water and P is the vapor pressure of the solution. It is already given to us that the vapor pressure of pure water is $25.21$ torr.
Now we establish a relation that the relative lowering of vapor pressure is equal to the mole fraction. By substituting the above values in the given relation, we get $\dfrac{{25.21 - P}}{{25.21}} = \dfrac{{0.111}}{{4.001}}$
On solving, we get $P = 24.5$ torr.
Therefore, the vapor pressure of the solution is $24.5$ torr.
Hence, option (D) is correct.
Note:It is to be noted that the Relative lowering of vapor pressure is the measure of the fall of vapor pressure of a solution when a solute is added to it. It is given by the formula $\dfrac{{{P^0} - P}}{{{P^0}}}$ where ${P^0}$ is the initial vapor pressure and P is the final vapor pressure.
Complete step by step solution:
Given to us that $20$ grams of glucose is dissolved in $70$ grams of water.
Let us first find the number of moles of glucose present in water. Now we know that the formula for number of moles is the ratio of weight of the solute and the gram molecular weight of the solute.
Hence, the number of moles of glucose is $\dfrac{{20}}{{180}} = 0.111$
Now, let us find the number of moles of water in similar fashion.
Number of moles of water is $\dfrac{{70}}{{18}} = 3.89$
Now, the formula for mole fraction of glucose is given as the ratio of number of moles of glucose to the total number of moles.
The number of moles of glucose is $0.111$
The total number of moles is $0.111 + 3.89 = 4.001$
The formula for relative lowering of vapor pressure is given as $\dfrac{{{P^0} - P}}{{{P^0}}}$
Here, ${P^0}$ is the vapor pressure of pure water and P is the vapor pressure of the solution. It is already given to us that the vapor pressure of pure water is $25.21$ torr.
Now we establish a relation that the relative lowering of vapor pressure is equal to the mole fraction. By substituting the above values in the given relation, we get $\dfrac{{25.21 - P}}{{25.21}} = \dfrac{{0.111}}{{4.001}}$
On solving, we get $P = 24.5$ torr.
Therefore, the vapor pressure of the solution is $24.5$ torr.
Hence, option (D) is correct.
Note:It is to be noted that the Relative lowering of vapor pressure is the measure of the fall of vapor pressure of a solution when a solute is added to it. It is given by the formula $\dfrac{{{P^0} - P}}{{{P^0}}}$ where ${P^0}$ is the initial vapor pressure and P is the final vapor pressure.
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