Question

How many grams of glucose ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ should be dissolved in 0.5 kg of water at 25 $^{o}C$ to reduce the vapor pressure of the water by 1.0 % ?
A. 50.5 g
B. 50.0 g
C. 18.0 g
D. 18.2 g

Answer 1

There is a formula to calculate the mole of glucose which is present in the 0.5 kg of water and it is as follows.
$\text{mole fraction of the solute =}\dfrac{n}{n+N}$
Here n = number of moles of the solute
N = number of moles of the water.

Complete Solution :
- In the question it is given that to calculate the number of grams of glucose is required to reduce the vapor pressure of the 0.5 kg of water by 1%.
- Mass of the glucose = 180.
- Assume the moles of glucose are ‘n’.
- Then number of moles water = $\dfrac{500g}{18}=27.77 moles$
- The number of moles of water in 0.5 kg of water N = 27.77 moles.
- The vapor pressure should be reduced by 1% means $\dfrac{0.01{{P}^{o}}}{{{P}^{o}}}$ .
- Here ${{P}^{o}}$ = Vapor pressure of the water.
- Therefore vapor pressure of the water is equal to number of moles of glucose, then
$\dfrac{n}{n+N}=\dfrac{0.01{{P}^{o}}}{{{P}^{o}}}$
- Now substitute N value in the above formula to get the number of moles of glucose.

& \dfrac{n}{n + N}=\dfrac{0.01{{P}^{o}}}{{{P}^{o}}} \\\ & \dfrac{n}{n + 27.7}=\dfrac{1}{100} \\\ & n=0.28moles \\\ \end{aligned}$$ \- The number of moles of glucose required to decrease the vapor pressure of the 0.5 kg of water is 0.28 moles. \- But we need the mass of the glucose required to reduce the vapor pressure of the 0.5 kg of water is 0.28 moles. \- Therefore the mass of the glucose required is = (0.28) (180) = 50.5 g. **So, the correct answer is “Option B”.** **Note:** Whenever we are going to add a non-volatile solute to water, the vapor pressure of the water is going to decrease. Because the non-volatile solute occupies the surface of the water and won’t allow the water to evaporate.