Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Text Solution 5M50M0.005M0.5M Answer : C Solution : In the given question 0.9 `gL^(-1)` means that 1000(mL) solution contains 0.9g of glucose. <br> `therefore` Number of moles `=0.9g "glucose"=(0.9)/(180)` mol glucose <br> `=5xx10^(-3)` mol glucose <br> (where , molecular mass of glucose `(C_(6)H_(12)O_(6))=12xx6+12xx1+6xx16=180u`) <br> i.e. L solution contains 0.05 mole glucose or the molarity of glucose is 0.005M. If the concentration of glucose \[\ce{(C6H12O6)}\] in blood is 0.9 g L–1, what will be the molarity of glucose in blood? 0.005 M Explanation: Molar mass of glucose \[\ce{(C6H12O6)}\] = 12 × 6 + 1 × 12 + 16 × 6 = 180 Molarity = `("conc. in gL"^-1)/"Molar mass" = (0.90 "gL"^-1)/(180 "g mol"^-1)` = 0.005 M Concept: Stoichiometry and Stoichiometric Calculations - Reactions in Solutions Is there an error in this question or solution? If the concentration of glucose (C6H12O6) in blood is 0.9 g L-1, what will be the molarity of glucose in blood? |