Matcn 0.5071 0.32 2.10 0.90.47 gl 2.112.1g12.1 912.10 919g19 gl 6.5816.58g16.58

Question

Matcn 0.5071 0.32 2.10 0.90.47 gl 2.112.1g12.1 912.10 919g19 gl 6.5816.58g16.58 916.59 Q: A pharmacist prepares the formulation below. How many grams of sodium chloride is added to the mixture? Drug A (E-value 0.32) 1 (E-value 0.21) 10 g Water ad. qs. qs. isotonic guided questions: -What is the sodium chloride equivalent grams of drug A?_19 (2 sig fig)-What is the sodium chloride equivalent grams of excipient 1?2g (3 sig fig) -How many grams of sodium chloride would be in a 1 L of a pure isotonic solution?3g-How many grams of sodium chloride should be added to the formulation?49 (3 sig fig) 1 g Excipient 1 L Sodium chloride ad. B1: 0.3210.32g10.32 gl B2: 2.112.1g12.1 g12.10 B3: 919g19 gl B4: 6.5816.58g16.58 916.59 Match 0.25/1 40140g/LI40 g/L 474.6 2 8 0.53710.537mol/L 0.537 mol/LI 2120sm/mol 2 Osm/moll 1.0711.07Osm/Ll 1.07 Osm/mol l Q: How many Osm/L are in a solution of 4 %w/v potassium chloride, KCl (m.w. 74.55)? guided questions:-Express 4%w/v as g/L?-1-g/L-How many mol/L of KCI?_2(3 sig fig) -What is the conversion between Osm and mol of KCI in Osm/mol?3 Osm/mol -How many Osm/L are in the solution?_4_Osm/L (3 sig fig) B1: 40140g/L140 g/L B2: 0.53710.537mol/LI0.537 mol/Ll B3: 2120sm/moll2 Osm/moll B4: 1.0711.070sm/L11.07 Osm/moll

Explanation / Answer

Sodium chloride present in drug is 0.9%

Drug A (E - value) = 0.32

Excipient E-value = 0.21

Assuming Drug A is 1 g in 1 L

So Amount of drug A = 1 * 0.32 = 0.32 g or 320 mg

Amount of NaCl = 0.9/100 X 1000 = 9 g

So grams of NaCl in drug A = 9 - 0.32 = 8.68 g

Grams of Excipient = 1 g

So Amount of drug A = 1 * 0.21 = 0.21 g or 210 mg

Amount of NaCl = 0.9/100 X 1000 = 9 g

So grams of NaCl in Excipient = 9 - 0.21 = 8.79 g

Amount of NaCl in 1L of pure isotonic solution = 0.9 X 100/ 1000 = 9 g

Total NaCl required in the formulation = 8.68 g

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