1) In this lab, you prepared 3.5% NaCl solution in water. In a typical intraveno

Question

1) In this lab, you prepared 3.5% NaCl solution in water. In a typical intravenous (IV) bag, the concentration is 0.9% NaCl (w/v) or 0.9% saline. This is to maintain the same osmotic pressure as that of blood plasma. Calculate how much NaCl (in grams) is needed to prepare a 24.0 L solution of saline. (Hint: 1g/mL = 100% w/v)
I would appreciate a step by step process, will rate asap thank you very much
1) In this lab, you prepared 3.5% NaCl solution in water. In a typical intravenous (IV) bag, the concentration is 0.9% NaCl (w/v) or 0.9% saline. This is to maintain the same osmotic pressure as that of blood plasma. Calculate how much NaCl (in grams) is needed to prepare a 24.0 L solution of saline. (Hint: 1g/mL = 100% w/v)
I would appreciate a step by step process, will rate asap thank you very much
1) In this lab, you prepared 3.5% NaCl solution in water. In a typical intravenous (IV) bag, the concentration is 0.9% NaCl (w/v) or 0.9% saline. This is to maintain the same osmotic pressure as that of blood plasma. Calculate how much NaCl (in grams) is needed to prepare a 24.0 L solution of saline. (Hint: 1g/mL = 100% w/v)
I would appreciate a step by step process, will rate asap thank you very much

Explanation / Answer

We know that

C1 x V1 = C2 x V2

Here,

C1 = 3.5 %
V1 = ?
C2 = 0.9 %
?V2 = 24 L

Substituting the values, we get;

(3.5%) x V1 = (0.9%) x (24 L)

or, V1 = (0.9%) x (24 L) / (3.5 %)

or, V1 = 6.17 L?

So, you take 6.17 L of 3.5% NaCl and add solvent (water) to make it 24 L solution, you will get 0.9% NaCl solution.

Now,

3.5% NaCl solution in water means

In 100 mL of of water, mass of NaCl = 3.5 g

or, In 1000 mL of of water, mass of NaCl = 35 g

or, In 1 L of of water, mass of NaCl = 35 g

or, In 6.17 L of of water, mass of NaCl = 6.17 x 35 g
?    = 215.95? g
= 216 g

Hence, you need 215.95 g of NaCl.

This question can be directly answered as

0.9 % NaCl solution in water means

In 100 mL of of water, mass of NaCl = 0.9 g

or, In 1000 mL of of water, mass of NaCl = 9 g

or, In 1 L of of water, mass of NaCl = 9 g

or, In 24 L of of water, mass of NaCl = 24 x 35 g
? = 216 g

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