3) You have the following stock solutions and solid chemicals in your lab: 1.5 M

Question

3) You have the following stock solutions and solid chemicals in your lab: 1.5 M NaCl 0.50 M Tris buffer, pH 8.0 25 mg/ml Bovine serum albumin (BSA) MgCl, (solid, anhydrous) MW = 95.211 g/mole (90% pure) Glucose (solid, anhydrous) How would you make 250 ml of a solution containing all of the following: (10 points) 0.1M NaCI, 0.05M Tris, pH 8.0, 0.25 mg/ml BSA, 10 mM MgCl 120 mg% glucose. List the volumes of each stock solution and the weights of each solid. Include calculations for each component.

Explanation / Answer

Answer:

1) Here we have to use M1V1=M2V2

Given M1=1.5 M NaCl, V1=?, M2=0.1 M and V2=250 mL

V1=M2V2/M1=(0.1Mx250 ml)/(1.5M)=16.66 ml.

Therefore 16.66 ml of 1.5M NaCl used to prepare 0.1 M NaCl of 2

50 ml solution.

2) Here also M1V1=M2V2

M1=0.5 M Tris Buffer, V1=?, M2=0.05 M, V2=250 ml

V1=(0.05 Mx250 ml)/0.5M=25 ml.

Therefore 25ml of 0.5 M Tris Buffer used to prepare 0.05M Tris Buffer of 250 ml.

3) Given M1=25 mg/ml BSA V1=?, M2=0.25 mg/ml and V2=250 ml.

(Here we can take M1 in mg/ml)

V1=(0.25mg/mlx250 ml)/(25mg/ml)

V1=2.5 ml.

Therefore 2.5 ml of 25 mg/ml BSA used to prepare 0.25 mg/ml BSA of 250 ml solution.

4) Given 10mM of MgCl=10x10^-3 M.

We know that molarity=(mass/molar mass)(1/Volume)

Molar mass of MgCl=95.21 g/mol and volume=250 ml=0.250 L.

10x10^-3 mol/L=(mass/95.21g/mol)(1/0.250 L)

Mass=0.238 g.

Therefore 0.238 g of MgCl used to prepare 10 mM MgCl of 250 ml.

5) 120 mg% Glucose,

We know that x% means X grams of solid dissolved in 100 ml of solution.

Therefore 120mg% means 120 mg in 100 ml of solution.

But we need to prepare 250 ml, then weight=(250 mlx120mg)/100 ml

=300 mg.

Therefore 300mg in 250 ml solution is equal to 120 mg% Glucose.

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