What is the mole fraction of 90 g of glucose (C₆H₁₂O6) dissolved in 900 g of water (H2O)?

0.009

Molar mass of C₆H₁₂O₆ = 12 × 6 + 1 × 12 + 16 × 6 = 180 g mol–1 .

Moles of C₆H₁₂O₆  = \(\frac{90}{180}\) = 0.5 mol

Moles of H₂O = \(\frac{900}{18}\) = 50 mol

mole fraction of C₆H₁₂O₆ = \(\frac{\text{moles of  } C_6H_{12}O_6 }{\text{moles of} C_6H_{12}O_6 +\text{ moles of }H_2O }\)

mole fraction of C₆H₁₂O₆ = \(\frac{0.5}{0.5+50}\)

mole fraction of C₆H₁₂O₆ = \(\frac{0.5}{50.5}\)

mole fraction of C₆H₁₂O₆ = 0.009