A solution containing 30 g of a non-volatile non-electrolyte solute in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) Molar mass of the solute, (ii) Vapour pressure of pure water at 298 K.

Question

Accepted Answer

Given Data: Initial mass of solute ($w2$) = $30	ext{ g}$ Initial mass of water ($w1$) = $90	ext{ g}$ Molar mass of water ($M1$) = $18	ext{ g mol}^{-1}$ Initial vapour pressure of solution ($p1$) = $2.8	ext{ kPa}$ Step 1: Set up the equation for the first case using Raoult's Law
Let the molar mass of the solute be $M2$ and the vapour pressure of pure water be $p^0$.
Number of moles of water initially ($n1$) = $\frac{90}{18} = 5	ext{ moles}$
Number of moles of solute ($n2$) = $\frac{30}{M2}$ 
According to Raoult's Law for relative lowering of vapour pressure: $$\frac{p^0 - p1}{p^0} = \frac{n2}{n1 + n2}$$ $$\frac{p^0 - 2.8}{p^0} = \frac{\frac{30}{M2}}{5 + \frac{30}{M2}} \quad 	ext{--- (Equation 1)}$$ Step 2: Set up the equation for the second case
When 18 g of water is added: New mass of water = $90 + 18 = 108	ext{ g}$ New number of moles of water ($n1'$) = $\frac{108}{18} = 6	ext{ moles}$ New vapour pressure of solution ($p2$) = $2.9	ext{ kPa}$ 
Applying Raoult's Law again: $$\frac{p^0 - 2.9}{p^0} = \frac{\frac{30}{M2}}{6 + \frac{30}{M2}} \quad 	ext{--- (Equation 2)}$$ Step 3: Solve Equation 1 and Equation 2
Divide Equation 1 by Equation 2: $$\frac{p^0 - 2.8}{p^0 - 2.9} = \frac{\frac{30}{M2} / (5 + \frac{30}{M2})}{\frac{30}{M2} / (6 + \frac{30}{M2})}$$ $$\frac{p^0 - 2.8}{p^0 - 2.9} = \frac{6 + \frac{30}{M2}}{5 + \frac{30}{M2}}$$ 
Alternatively, we can express relative lowering as: $$\frac{2.8}{p^0} = 1 - \frac{30 / M2}{5 + 30/M2} = \frac{5}{5 + 30/M2}$$
Similarly for second case: $$\frac{2.9}{p^0} = \frac{6}{6 + 30/M2}$$ 
Dividing both equations: $$\frac{2.8}{2.9} = \frac{5 / (5 + 30/M2)}{6 / (6 + 30/M2)}$$
Let $x = \frac{30}{M2}$. $$\frac{2.8}{2.9} = \frac{5(6 + x)}{6(5 + x)}$$ $$2.8 	imes 6(5 + x) = 2.9 	imes 5(6 + x)$$ $$16.8(5 + x) = 14.5(6 + x)$$ $$84 + 16.8x = 87 + 14.5x$$ $$16.8x - 14.5x = 87 - 84$$ $$2.3x = 3$$ $$x = \frac{3}{2.3} = 1.304$$ 
Since $x = \frac{30}{M2} = 1.304$: $$M2 = \frac{30}{1.304} = 23.0	ext{ g mol}^{-1}	ext{ (Molar mass of solute)}$$ Step 4: Calculate Vapour Pressure of Pure Water ($p^0$)
Substitute the value of $x$ back into Equation 1: $$\frac{p^0 - 2.8}{p^0} = \frac{1.304}{5 + 1.304} = \frac{1.304}{6.304} = 0.2068$$ $$p^0 - 2.8 = 0.2068 p^0$$ $$p^0 - 0.2068 p^0 = 2.8$$ $$0.7932 p^0 = 2.8$$ $$p^0 = \frac{2.8}{0.7932} = 3.53	ext{ kPa}$$ Final Answer: (i) Molar mass of the solute = $23.0	ext{ g mol}^{-1}$ (ii) Vapour pressure of pure water = $3.53	ext{ kPa}$

Given Data:

Step 1: Set up the equation for the first case using Raoult's Law\nLet the molar mass of the solute be $M_2$ and the vapour pressure of pure water be $p^0$.\nNumber of moles of water initially ($n_1$) = $\frac{90}{18} = 5\text{ moles}$\nNumber of moles of solute ($n_2$) = $\frac{30}{M_2}$

Step 2: Set up the equation for the second case\nWhen 18 g of water is added:

Step 3: Solve Equation 1 and Equation 2\nDivide Equation 1 by Equation 2:

Step 4: Calculate Vapour Pressure of Pure Water ($p^0$)\nSubstitute the value of $x$ back into Equation 1:

Final Answer: