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nadya68 [22]
2 years ago
4

During the rGFP purification experiment, the instructor will have to make breaking buffer for the students to use. This buffer c

ontains 150mM NaCl. Given a bottle of crystalline NaCl (M.W. = 40g/mole), describe how you would make 500ml of 150mM NaCl. Include the type and sizes of any measuring vessels that you use to make the solution.
Chemistry
1 answer:
Leya [2.2K]2 years ago
6 0

Answer:

dealing with a buffer

use the formula C = n/V...........1

therefore say n = m/Mr.........2

then substitute equation 2 into equation 1

∴ CV = m/Mr

then m = CVMr

Since we are looking for mass of  Nacl to make a buffer solution.

m = 150×10^{-3} ml^{-1} × 500×10^{-3} l × 40 g/mol

mass of Nacl = 3g

Explanation: You first weight out 3g of 150mM of NaCl and then dissolve it into a 1000ml beaker dilute it using a  volume of 500ml of water to make volume.

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The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

                       In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

                             Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

                    HCOOH (formic acid)    →     HCOO⁻ (formate)  +  H⁺

                H₃CCOOH (acetic acid)     →      H₃CCOO⁻ (acetate)  +  H⁺

Formal Charges:

                           Formal charges are calculated using following formula,

          F.C  =  [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

                    F.C  =  [6] - [6 + 2/2]

                    F.C  =  [6] - [6 + 1]

                    F.C  =  6 - 7

                    F.C  =  -1

For Sodium:

                    F.C  =  [1] - [0 + 0/2]

                    F.C  =  [1] - [0]

                    F.C  =  1 - 0

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