1. **State the problem:** We are given the vapor pressure of pure water at 20°C as 17.5 atm. When 114 g of sucrose is dissolved in 1000 g of water, the vapor pressure decreases by 0.11 atm. We need to find the molecular mass ($M$) of the sucrose. 2. **Write known values:** - Vapor pressure of pure water, $P_0 = 17.5$ atm - Vapor pressure of solution, $P = 17.5 - 0.11 = 17.39$ atm - Mass of sucrose, $m_2 = 114$ g - Mass of water, $m_1 = 1000$ g - Molar mass of water, $M_1 = 18$ g/mol (known) 3. **Apply Raoult's law:** The lowering of vapor pressure is proportional to the mole fraction of solute. $$\Delta P = P_0 - P = P_0 \times x_2$$ where $x_2$ is the mole fraction of the solute. 4. **Calculate mole fraction of sucrose $x_2$:** $$x_2 = \frac{\Delta P}{P_0} = \frac{0.11}{17.5} = 0.0062857$$ 5. **Express mole fraction $x_2$ in terms of moles:** $$x_2 = \frac{n_2}{n_1 + n_2}$$ Since sucrose is nonvolatile and usually $n_2 \ll n_1$, approximation: $$x_2 \approx \frac{n_2}{n_1}$$ 6. **Find moles of water $n_1$:** $$n_1 = \frac{m_1}{M_1} = \frac{1000}{18} = 55.56 \, \text{mol}$$ 7. **Find moles of sucrose $n_2$:** $$n_2 = x_2 \times n_1 = 0.0062857 \times 55.56 = 0.3495 \, \text{mol}$$ 8. **Calculate molecular mass $M$ of sucrose:** $$M = \frac{m_2}{n_2} = \frac{114}{0.3495} = 326.3 \, \text{g/mol}$$ **Final answer:** The molecular mass of sucrose is approximately $326.3$ g/mol.