Physicain elects to re-hydrate the patient br intravenous(IV) administration of 2.0 L of 0.45% (w/v) of NaCl. 0.45%9w/v) = 0.45g/100ml
A.How many grams of NaCl were given?
B.How many moles of Nacl were given?
C.What is the molarity of the NaCl solution?
0.45 g NaCl x (2000/100) = ?g NaCl in 2 L.
mols NaCl = grams NaCl/molar mass NaCl.
M NaCl = mols NaCl/2L
To solve these questions, we need to use the given information about the concentration of the solution and the volume administered. Let's break it down step by step:
A. To find the grams of NaCl given:
Given concentration of NaCl solution = 0.45% (w/v)
0.45% (w/v) = 0.45 g/100 ml
Given volume administered = 2.0 L = 2000 ml (since 1 L = 1000 ml)
To find the grams of NaCl given, we can use the proportion:
(0.45 g/100 ml) = (x g/2000 ml)
Cross-multiplying and solving for 'x' gives us:
x = (0.45 g/100 ml) × 2000 ml
x = 9 g
Therefore, 9 grams of NaCl were given.
B. To find the moles of NaCl given:
To find the moles, we need to use the molar mass of NaCl, which is approximately 58.5 g/mol.
Moles of NaCl = (grams of NaCl given) / (molar mass of NaCl)
Moles of NaCl = 9 g / 58.5 g/mol
Moles of NaCl ≈ 0.154 mol
Therefore, approximately 0.154 moles of NaCl were given.
C. To find the molarity of the NaCl solution:
Molarity = (moles of solute) / (volume of solution in liters)
Given moles of NaCl = 0.154 mol
Given volume of solution = 2.0 L
Molarity = 0.154 mol / 2.0 L
Molarity ≈ 0.077 M
Therefore, the molarity of the NaCl solution is approximately 0.077 M.