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1. Determine the quantity of dextrose and/or sodium chloride in each one of the following IV solutions. a. 250 ml salt sodium chloride

0.9 g Xg × 100 ml 250 ml 100 x 225 = = 2.25 g 100 100

2.25g

b. 3 l of D5NS dextrose sodium chloride

5g Xg × 100 ml 3000 ml 100 x 15000 = = 150 100 100

150g 27g

0.9 g Xg × 100 ml 3000 ml 100 x 2700 = = 27 100 100

c. 500 ml of D5 0.33% NaCl dextrose sodium chloride

5g Xg × 100 ml 500 ml 100 x 2500 = = 25 100 100

25g 1.65g

0.33 g Xg × 100 ml 500 ml 100 x 165 = = 1.65 100 100

d. 250 ml of D10W dextrose

10 g Xg × 100 ml 250 ml 100 x 2500 = = 25 100 100

25g

D10

e. 1,000 ml of D5NS

dextrose sodium chloride

5g Xg × 100 ml 1000 ml 100 x 5000 = = 50 100 100

50g 9g

0.9 g Xg × 100 ml 1000 ml 100 x 900 = =9 100 100

f. 750 ml of NS sodium chloride

.9 g Xg × 100 ml 750 ml 100 x 675 = = 6.75 100 100

6.75g

g. 0.75 L of 0.45% NaCl NaCl 3.375 g

0.45 g Xg × 100 ml 750 ml .75l = 750ml 100 x 337 .5 = = 3.375 100 100

g

h. 300 ml of D12 0.9% salt dextrose sodium chloride

12 g Xg × 100 ml 300 ml 100 x 3600 = = 36 100 100

36g 2.7g

0.9 g Xg × 100 ml 300 ml 100 x 270 = = 2.7 100 100

i. 500 ml of D51/2 NS dextrose sodium chloride 25g 2.25g

5g Xg × 100 ml 500 ml 100 x 2500 = = 25 100 100

1 ÷ 2 × 0.9 = 0.45 0.45 g Xg × 100 ml 500 ml 100 x 225 = = 2.25 100 100

j. 0.5 l of D101/4 NS dextrose sodium chloride 50g 1.125g

D10

10 g Xg × 100 ml 500 ml 100 x 5000 = = 50 100 100

1 ÷ 4 × 0.9 = 0.225 0.225 g Xg × 100 ml 500 ml 100 x 112 .5 = =1.125 100 100

For the following questions, make use of the drop factor to determine the flow rate in gtt / min. Ordered: 1l of 0.45% salt IV at 200 ml per hr

1. Drop factor 10 gtt per ml

Flow rate:

33

gtt per min

2. Drop factor 15 gtt per ml

Flow rate:

50gtt / min

3. Drop factor 20 gtt / ml

Flow…...

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