If a patient consumes 2.50 L of a 5.00% glucose solution (d = 1.10 g/mL) during

Question

If a patient consumes 2.50 L of a 5.00% glucose solution (d = 1.10 g/mL) during a 1 day period, how many grams of glucose (MM = 180) has the patient consumed? How many grams of NaBr must be added to 77.5 g of water in order to prepare a 25.0 wt% solution of NaBr? 0.75 mol of Ca(OH)_2 contains how many equivalents? What is the equivalent weight of barium hydroxide (MM = 171)? What is the normality of a solution prepared by mixing 9.50 g of barium hydroxide (MM = 171) in enough water to make 2000. mL of solution?

Explanation / Answer

22)

volume = 2.50 L

Mass % = 5.00 %

molarity = mass % x density x 10 / Molar mass

              = 5 x 1.10 x 10 / 180

              = 0.306 M

Molarity = moles / volume

0.306 = moles / 2.50

moles = 0.764

mass of glucoe = 0.764 x 180 = 137.5 g

mass of glucose = 137.5 g

23)

weight % = (weight of solute / weight of solution ) x 100

25 = (x / 77.5 + x ) x 100

x = 25.83

mass of NaBr = 25.8 g

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