Question

The vapour pressure of water at 23 degree celsius is 19.8 mm 0.1 mole of glucose is dissolved in 178.2 grams of water what is the vapour pressure of resulting solution at same temperature

Answer 1

Answer:

The answer is 19.602 mm

Explanation:

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Answer 2

The vapor pressure of the solution at $23^0C$ is 19.6 mm 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)

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

moles of solute (glucose) = 0.1

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 = 10.0

$x_2$ = mole fraction of solute =$\frac{0.1}{10.0}=0.01$

$\frac{19.8-p_s}{19.8}=1\times 0.01$

$p_s=19.6mmHg$

Thus the vapor pressure of the solution at $23^0C$ is 19.6 mm Hg

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