Question

What is the change in vapor pressure when 74.10 g fructose, C6H12O6, are added to 181.5 g water (H2O) at 298 K (vapor pressure of pure water at 298 K = 3.1690 kPa, molar mass of fructose = 180.156 g/mol, molar mass of water = 18.02 g/mol)?

Answer 1

The change in vapor pressure when 74.10 g fructose, C6H12O6, are added to 181.5 g water is 0.146.

What is vapor pressure?

Vapor pressure is the  measure or tendency of  a substance to change into a vapor or gaseous state.

$\rm Mole\; fraction\; of\; water = \dfrac{nH_2O}{nH_2O+npent.}$

Calculate the moles of water and fructose

$\rm Number\;of \;moles= \dfrac{181.5 g}{18.02}= 10.072\\\\\rm Number\;of \;moles= \dfrac{74.10 g }{180.156 g/mol}=0.4113$

The total number of moles are

$\dfrac{10.072}{10.072 + 0.486} = \dfrac{10.072}{10.558} = 0.953$

Calculating the vapor pressure

0.953 x 3.1690 kPa = 3.023

Change in vapor pressure

3.1690 kPa - 3.023 = 0.146

Thus, the  change in vapor pressure is 0.146.

Learn more about vapor pressure

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