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Determine the vapor pressure of a solution at 25°C that contains 76.6 g of glucose (C6H12O6 ) in 250.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr

User Edur
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1 Answer

6 votes

Answer:


23.093\ \text{torr}

Step-by-step explanation:


M_g = Molar mass of glucose = 180.2 g/mol


M_w = Molar mass of water = 18 g/mol


m_g = Mass of glucose = 76.6 g


m_w = Mass of water =
250* 1\ \text{g/mL}=250\text{g}


P_0 = Vapor pressure of pure water at 25°C = 23.8 torr

The mole fraction of glucose is


x_g=((m_g)/(M_g))/((m_g)/(M_g)+(m_w)/(M_w))=((76.6)/(180.2))/((76.6)/(180.2)+(250)/(18))\\\Rightarrow x_g=0.0297

Mole fraction of the solute would be


(P_0-P)/(P_0)=x_g\\\Rightarrow 0.0297=(23.8-P)/(23.8)\\\Rightarrow P=23.8-0.0297*23.8\\\Rightarrow P=23.093\ \text{torr}

The vapor pressure of the solution is
23.093\ \text{torr}.

User Flignats
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