The total concentration of disssolvedparticles in blood is 0.30 M . An intraveno

Question

The total concentration of disssolvedparticles in blood is 0.30 M. An intravenous (IV) solutionmust be isotonic with with blood, which means it must have the sameconcentration. (a) To relievedehydration, a patient is given 135mL/h of IV glucose (C6H12O6)for 2.0 h. What mass (g) of glucosedid she receive?

(b) If isotonic saline (NaCl) were used, what is the molarity ofthe solution?

(c) If the patient is given 175 mL/hof IV saline for 2.5 h, how many gramsof NaCl did she receive?

I did part (a)
135mL/h * 2h = 270mL * 1L/1000mL = 0.270 L * 0.3mol/L =0.081 mol
C6H12O6=180g/mol * 0.081mol =14.58g
but I'm not sure how to do parts (b) and (c)



(a) To relievedehydration, a patient is given 135mL/h of IV glucose (C6H12O6)for 2.0 h. What mass (g) of glucosedid she receive?

(b) If isotonic saline (NaCl) were used, what is the molarity ofthe solution?

(c) If the patient is given 175 mL/hof IV saline for 2.5 h, how many gramsof NaCl did she receive?

I did part (a)
135mL/h * 2h = 270mL * 1L/1000mL = 0.270 L * 0.3mol/L =0.081 mol
C6H12O6=180g/mol * 0.081mol =14.58g
but I'm not sure how to do parts (b) and (c)

Explanation / Answer

(a) In 2 hour, the patient recieve 135 x 2 = 270mL of IV glucose c(glucose) = n/v n(Glucose) = cV = 0.30 x 0.27 = 0.081mol n(Glucose) = m/M m(Glucose) = nM = 0.081 x 180.156 = 14.59g (b) Molarity of the solution must be 0.3M as well (must have thesame concentration) (c) In 2.5 hr, the patient receive 175 x 2.5 = 437.5mL of NaCl c(NaCl) = n/v n(NaCl) = cV = 0.3 x 0.4375 = 0.13125mol n(NaCl) = m/M m(NaCl) = nM = 0.13125 x 58.44 = 7.67g Hope this helps!

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