Suppose 20.23 g of glucose are dissolved in 990 g of water. Glucose is nonvolati

Question

Suppose 20.23 g of glucose are dissolved in 990 g of water. Glucose is nonvolatile (has no vapor pressure) and has molecular 180.2 g/mole. Water has a vapor wt of and a pure pressure (PA degree) 26.7 mm Hg (at 27 degree C) molecular of weight g/mole. Find the moles of each component. moles of A Then, determinee the mole fractions (x_Glucose and X_water of each component using xA total moles What should X_Glucose + X_water Finally, use Raoult's Law to determinee the total va pressure of the solution: P_A = (PA degree) (X_A)

Explanation / Answer

m = 20.23 g of glucose

m = 99 g of water

MW = 180.2 g/mol

P° = 26.7 mm Hg

a)

mole fractions:

mol of glucose = mass/MW = 23.23/180.2 = 0.12891

mol of water = mass/MW = 99/18 = 5.5

total mol = 5.5+0.12891 = 5.62891

so

x-water= 5.5/5.62891 = 0.977098

x-glucose = 0.12891/5.62891 = 0.022905

b)

so

xglucose + xwater = 1 always, since ther are only two species

c)

Raoult law:

Pmix = P°*Xa

dPmix = 26.7*0.022905

Ptotal = 0.61156

Pmix = 26.7-0.61156 = 26.08844 mm Hg for mix

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