Solution:
Mass percent concentration at 25% means that for each 100g of solution, mass of solute is 25g and mass of solvent is 75 g (1:3 ratio)
"Molality=\\frac{moles\\,of\\, solvent}{1\\,kg\\,of\\,solute}"
Mass of glucose that corresponds to 1 kg of solute is "\\frac{1\\,kg}{3}=333.3g"
Moles of glucose: "\\frac{333.3g}{180.156 g\/mol}=1.85 mol"
Molality=1.85 mol/kg
Moles of 1kg of water: "\\frac{1000g}{18.015 g\/mol}=55.51 mol"
"Mole\\,fraction=\\frac{moles\\,of\\,solvent}{moles\\,of\\,solvent+moles\\,of\\,solute}"
"Mole\\,fraction=\\frac{1.85\\,mol}{1.85\\,mol+55.51\\,mol}=0.032"
Mole fraction=0.032
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