Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: WuwuAndNillump on December 13, 2019, 02:18:00 AM
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The situation basically starts like this, and I've been going about it for 3 nights to no avail.
SITUATION:
Dextrose is a form of sugar (mixture of glucose, C5H12O6, and water) and is usually injected into a patient through intravenous infusion. It is used to treat dehydration, low blood sugar, and insulin shock.
In disasters, especially during earthquakes, typhoons, and tsunamis, it is very common to have lacking supplies in medicines to treat mass injured patients with lack of means of transportation in delivering fresh supplies.
1. You need to administer 5.00% mass-volume dextrose to a patient; however, the available concentrations are not expressed in mass-volumes, which of the following can be a substitute? Density of glucose = 1.56 g/mL. Density of water = 1g/mL.
Available substitute concentrations:
a. 0.298 M
b. 5.50x10-3 mole fraction glucose
CHOICES:
I. 0.298 M, because they have the same ratio of the amount of glucose and amount of solution
Ii. 5.50 x10-3 mole fraction glucose, because they have the same ratio of the amount glucose and amount of solution
Iii. both i and ii can be substitutes since they they have equivalent ratio of solute and solution with the 5 % mass-volume dextrose
Iv. not enough data is provided
EDIT: I've tried putting hypothetical volumes of dextrose using the given mass-percent, then used it to calculate Molarity, but I'm always a bit off.
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The percentage concentration for both solutions can be easily calculated. For a molar concentration solution, you must assume a density equal to the density of water or find the exact value in the tables (~ 1.02).
Of course, both solutions do not have exactly 5% concentration, so some acceptable error must be assumed (for medical concentrations, an error below 10% is usually acceptable).
Solid glucose density is useless.
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The percentage concentration for both solutions can be easily calculated. For a molar concentration solution, you must assume a density equal to the density of water or find the exact value in the tables (~ 1.02).
Of course, both solutions do not have exactly 5% concentration, so some acceptable error must be assumed (for medical concentrations, an error below 10% is usually acceptable).
Solid glucose density is useless.
I tried your suggestion and I got somewhere between 0.299 M and 0.300 M but never 0.298 M, is that still acceptable or is that point difference too far off to not be a plausible choice?
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You should convert to a percentage:
I. molar concentration; or II. mole fraction
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Wow. 3 nights is a long time. Let's see if I can give you a hand.
first off let's verify you have these definitions
molarity (M) = moles solute / Liter solution
molality (m) = moles solute / kg solvent
mole fraction in liquid phase (χ) = moles of 1 component / total moles
% mass / Volume in the medical field = g / 100mL
that last one is a bit ambiguous so let's make sure we're on the same page
5.00% m/V = 5.00g / 100mL solution
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now you have an issue with the problem statement..
the problem statement says
(1) dextrose (C5H12O6)
(2) density of glucose = 1.56 g/mL
(3) 0.298M
(4) 5.50x10^-3 mole fraction glucose
(I) 0.298M because they have the same ratio of glucose and volume solution
(II) 5.50x10^-3 mole fraction glucose because it has the same ratio of glucose and volume solution
(III) both I and II because they have the same ratio of solute and solvent as the 5.00% dextrose
notice a couple of things
(A) dextrose and glucose are used interchangeably
(B) dextrose has the formula C5H6O12.. that is INCORRECT
let's spend a second and review "dextrose" vs "glucose"
glucose is the molecule C6H12O6.. C6.. not C5
glucose exists as "d" and "l" enantiomers... (dextro and levo)
the "d-glucose" aka "dextro-glucose" is commonly called "dextrose"
meaing
the formula for dextrose is C6H12O6 with molar mass 180.16g/mol and the terms glucose and
dextrose are used interchangeably for this question. C5H12O6 has molar mass = 168.15 g/mol fyi
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let's do a quick conversion
0.298 mol dextrose 180.16g dextrose 53.7g dextrose 1 L 5.37g
---------------------- x -------------------- = ----------------- x ---------- = -------- = 5.37% (m/V)
1 L solution 1 mol dextrose 1L solution 1000mL 100mL
if we use the WRONG formula and wrong molar mass of dextrose, we get this calculation
0.298 mol dextrose 168.15g dextrose 50.1g dextrose 1 L 5.01g
---------------------- x -------------------- = ----------------- x ---------- = -------- = 5.01% (m/V)
1 L solution 1 mol dextrose 1L solution 1000mL 100mL
and we can see that with the wrong information, 0.298M glucose = 5.01% (m/V) dextrose
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Moving right along let's convert 5.50x10^-03 mole fraction glucose to % (m/V) glucose. let's start by assuming a total # of moles of say... 1000
mole dextrose moles dextrose
χ dextrose = ----------------- = ------------------ = 5.50x10^-3
(total moles) 1000mol
so that moles dextrose = 1000 mol * 5.50x10^-3 = 5.50
and mole water = 1000 mol - 5.50 mol = 994.5 mol H2O
next, let's calculate volume of each and then make the terrible assumption that the volumes are additive (which they aren't but it's an approximately starting point and seems to be where you're problem is leading you)
using the CORRECT molar mass for dextrose
5.50mol 180.16g 1mL
volume dextrose = ---------- x ---------- x ------- = 635.2mL
1 1 mol 1.56g
using the INCORRECT molar mass for dextrose
5.50mol 168.15g 1mL
volume dextrose = ---------- x ---------- x ------- = 592.8mL
1 1 mol 1.56g
The volume of water is
994.5mol 18.02g 1mL
volume water = ----------- x ---------- x ------- = 17921mL
1 1 mol 1g
total volume = 18556mL (correct mw for dextrose) or 18514mL (incorrect mw for dextrose)
giving this % m/V of dextrose using CORRECT mw of dextrose
5.50mol 180.16g 100 (5.50 / 18556 * 180.16 * 100) g dextrose
------------ x --------- x ---- = ---------------------------------------------- = 5.34% (m/V)
18556mL 1 mol 100 100mL solution
and this % m/V of dextrose using INCORRECT mw of dextrose
5.50mol 168.15g 100 (5.50 / 18514 * 168.15 * 100) g dextrose
------------ x --------- x ---- = ---------------------------------------------- = 5.00% (m/V)
18514mL 1 mol 100 100mL solution
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meaning...
.. IF we use the wrong formula for dextrose.... choice III is the correct choice
.. but if we use the correct formula for dextrose... none of the choices are correct.