Question

The vapour pressure of water at 20° C is 17.5 mm Hg. If 18 g of glucose (C₆H₁₂O₆) is added to 178.2 g of water at 20° C, the vapour pressure of the resulting solution will be (a) 17.325 mm Hg (b) 15.750 mm Hg(c) 16.500 mm Hg (d) 17.500 mm Hg

Answer 1

(a) : In solution containing nonvolatile solute,pressure is directly proportional to its mole fraction.

Psolution = vapour pressure of its pure component × mole fraction in solution

Psolution = 17.32.

hope this helps u.

if it really helps u plz mark e as a brain list

Answer 2

The vapor pressure of the resulting solution at $20^0C$ is 17.325mm Hg

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes like glucose)

$x_2$ = mole fraction of solute  =$\frac{\text {moles of solute}}{\text {total moles}}$

Given : 18 g of glucose is present in 178.2 g of water

moles of solute (glycerol) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{18g}{180g/mol}=0.1moles$

moles of solvent (water) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{178.2g}{18g/mol}=9.9moles$

Total moles = moles of solute (glucose)  + moles of solvent (water) = 0.1 + 9.9 = 10

$x_2$ = mole fraction of solute  =$\frac{0.1}{10}=0.01$

$\frac{17.5-p_s}{17.5}=1\times 0.01$

$p_s=17.325mm Hg$

Learn more about relative lowering of vapor pressure

https://brainly.in/question/10016774

https://brainly.com/question/12778562