Question 4.
(i) The weight of mannitol = 20 g/100 mL * 1000 mL = 200 g
The NaCl equilvent of mannitol = 200 g * 0.18 = 36 g
The actual amount of NaCl required = 0.9 g/100 mL * 1000 mL = 9 g
Here, the amount of mannitol (36 g) > required amount of NaCl (9 g), hence it is hypertonic.
(ii) The NaCl equivalent of monobasic sodium phosphate = 19 g * 0.49 = 9.31 g
The NaCl equivalent of dibasic sodium phosphate.7H2O = 7 g * 0.29 = 2.03 g
The total mass = 9.31 + 2.03 = 11.34 g
The actual amount of NaCl required = 0.9 g/100 mL * 120 mL = 1.08 g
Here also, the the amount of mixture of phosphates (11.34 g) > required amount of NaCl (1.08 g), hence it is hypertonic.
4. Determine if the following commercial products are hypotonic, isotonic, or hypertonic: 0 1 L of...
Determine whether the following commercial product is hypotonic, isotonic, or hypertonic. An ophthalmic solution containing 40 mg/mL of cromolyn sodium and 0.01% of benzalkonium chloride in purified water. Cromolyn sodium E value= 0.11 Benzalkonium chloride E value = 0.16
11 Isotonic and Buffer Solutions 193 CHLORIDE EQUIVALENTS (E-VALUES) (Continuedh Molecular Weight lons(E-valuer Sodium Chloride Equivalent 92 0.35 0.17 354 0.09 182 759 0.18 aH0 0.11 0.13 0.27 0.22 0.18 0.24 247 372 G potassium 0.32 perylephrine hydrochloride 413 245 74.5 166 101 136 273 331 438 170 0.16 0.24 0.76 salicylate 0.58 0.43 0.21 phosphate, monobasic 0.15 0.12 0.33 0.65 0.42 0.6 acaine hydrochloride hydrochloride Scopolamine hydrobromide 3H.O er nrate Sodium bicarbonate Sodum borate-10H,0 Sadum carbonate-H0 dum chloride Sdum...
#4.. please show work and dont skip steps. trying to learn how
to do this for the actual biochemistry lab
PART B. PRACTICE CALCULATIONS Convert units (show your work!): a. 0.01 ml b. 0.05 M = MM c. 0.2 M = mM d. 10 mM - M 2. Molar mass of Na olar mass of NaOH is 40 /mol. How many grams of NaOH do you have to take to prepare a. 11 of 10 M solution of NaOH? b....