The vapour pressure of water at 23° C is i9.8 mm of Hg.0.1 mol of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm Hg) of the resultant solution?

Question

The vapour pressure of water at 23° C is i9.8 mm of Hg.0.1 mol of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm Hg) of the resultant solution? (A) 19.0 (B) 19.602 (C) 19.402 (D) 19.202

Tardigrade Admin · Accepted Answer

Moles of glucose; nC6H12O6=0.1 Moles of water; nH2O=(178.2/18)=9.9 Mole fraction of water, xH2O=(9.9/9.9+0.1)=0.99 Psolution=Psolvent× xsolvent =19.8× 0.99 =19.602mm text Hg

Q. The vapour pressure of water at $23^{\circ} C$ is i9.8 mm of Hg.0.1 mol of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm Hg) of the resultant solution?

19.0

19.602

19.402

19.202

Moles of glucose; $n C_{6} H_{12} O_{6} = 0.1$ Moles of water; $n_{H_{2} O} = \frac{178.2}{18} = 9.9$ Mole fraction of water, $x_{H_{2} O} = \frac{9.9}{9.9 + 0.1} = 0.99$ $P_{so l u t i o n} = P_{so l v e n t} \times x_{so l v e n t}$ $= 19.8 \times 0.99$ $= 19.602 mm H g$

Q. The vapour pressure of water at 23∘C is i9.8 mm of Hg.0.1 mol of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm Hg) of the resultant solution?