A (0,1 M)
1.Weigh 0,8766 g ultrapure NaCl dried at 110 0C for 30 min.
2. Transfer NaCl in a clean 150 mL volumetric flask using a funnel. 3. Wash the funnel with 120 mL demineralized water.
4. Put the flask in a thermostat and maintain 30 min at 20 0C.
5. Add demineralized water up to the mark.
6. Stir vigorously and transfer in a clean bottle with stopper.
7. Add a label with necessary information.B (0,2 M)
1.Weigh 1, 7532 g ultrapure NaCl dried at 110 0C for 30 min.
2. Transfer NaCl in a clean 150 mL volumetric flask using a funnel. 3. Wash the funnel with 120 mL demineralized water.
4. Put the flask in a thermostat and maintain 30 min at 20 0C.
5. Add demineralized water up to the mark.
6. Stir vigorously and transfer in a clean bottle with stopper.
7. Add a label with necessary information.
You must use 8 part water and 2 part NaCl. For example, take 20cm qubed of NaCl and 80cm qubed of water.
Take 5.8g of NaCl.Then dissolve in 1dm3 of water.
Dissolve 8,766 g ultrapure, dried NaCl in demineralized water, at 20 0C, in a 1 L volumetric flask.
Dissolve 11,688 g NaCl (reagent grade) in 200 mL demineralized water at
20 0C, in a volumetric flask.
Dissolve 8,766 g NaCl reagent grade in 1 L demineralized water, at 20 0C in a volumetric flask.
A water solution containing 50 mM tris(hydroxymethyl)aminomethane and 150 mM sodium chloride has a pH of 7,6.
Also 150 mM of sodium.
The normal concentration of sodium in the blood plasma is 136-145 mM
The most common cation in the interstitial fluid is sodium at 150 mM. Next is calcium at 8.4 mM and potassium at 5 mM.
The needed mass is 0,0584 g.
The formula for this conversion is mEq = mg/atomic weight * valence. The atomic weight of sodium chloride is 23mg/mM.
Calculate the percent by weight of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl) * Calculate the molecular mass (MM):MM = 22.99 + 35.45 = 58.44 * Calculate the total mass of Na present:1 Na is present in the formula, mass = 22.99 * Calculate the percent by weight of Na in NaCl:%Na = (mass Na ÷ MM) x 100 = (22.99 ÷ 58.44) x 100 = 39.34% * Calculate the total mass of Cl present:1 Cl is present in the formula, mass = 35.45 * Calculate the percent by weight of Cl in NaCl:%Cl = (mass Cl ÷ MM) x 100 = (35.45 ÷ 58.44) x 100 = 60.66% The answers above are probably correct if %Na + %Cl = 100, that is,39.34 + 60.66 = 100.
100 mM means 0.1 M , So weight out 0.1 x 95.211 g = 9.52 g
150 mm, there is 10 mm in 1cm, therefore 150 mm would be equal 15 cm.
1 in = 25.4 mm. (150 mm)/(25.4 mm/in) = 5.9055 inches.
150 mm = 15 cm
Tartaric acid - 18.75 mM Sodium Tartrate Dihydrate - 6.25 mM pH of this mixture should be near 2.9, adjust pH down to 2.5 using HCl, and QS to 1 L by water.