A100.5 ml intraveneous (iv) solution contains 5.10 g glucose (c6h12o6). what is the molarity of this solution?

Answer is: the molarity of this glucose solution is 0.278 M.
m(C₆H₁₂O₆) = 5.10 g.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆) .
n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆) = 0.028 mol.
V(solution) = 100.5 mL ÷ 1000 mL/L.
V(solution) = 0.1005 L.
c(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ V(solution).
c(C₆H₁₂O₆) = 0.028 mol ÷ 0.1005 L.
c(C₆H₁₂O₆) = 0.278 mol/L.