A 100.5 ml intraveneous (iv) solution contains 5.10 g glucose (c6h12o6). what is the molarity of this solution?
1 answer:
Answer is: <span>the molarity of this glucose solution is 0.278 M. m</span>(C₆H₁₂O₆<span>) = 5.10 g. n</span>(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆<span>) . </span>n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol. n(C₆H₁₂O₆<span>) = 0.028 mol. </span>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<span>) = 0.278 mol/L.</span>
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