Help Solving Percent Molarity Problem

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SUMMARY

The discussion focuses on preparing a 200 mL solution of 2.5% Glutaraldehyde from a 25% stock solution in a 0.2M Sodium Cacodylate Buffer. The molecular weight of Glutaraldehyde is established as 100 g/mol, and Sodium Cacodylate is 160 g/mol. The user correctly calculates that the 25% stock solution contains 250 g of Glutaraldehyde, equating to a concentration of 2.5M. The key formula to use for dilution is M1V1 = M2V2, which allows for the determination of the required volume of the stock solution needed to achieve the desired concentration.

PREREQUISITES
  • Understanding of molarity and percent concentration
  • Familiarity with the dilution formula M1V1 = M2V2
  • Knowledge of molecular weight calculations
  • Basic laboratory skills for preparing solutions
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  • Practice using the dilution formula M1V1 = M2V2 with different concentrations
  • Learn about preparing solutions with varying percent concentrations
  • Explore the properties and uses of Glutaraldehyde in laboratory settings
  • Investigate the role of Sodium Cacodylate Buffer in biochemical applications
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Chemistry students, laboratory technicians, and researchers involved in solution preparation and concentration calculations.

depeche
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I have to prepare the following solution:

200 mL of 2.5% Glutaraldehyde from 25% stock Glutaraldehyde in 0.2M Sodium Cacodylate Buffer

So far I know:

Molecular weight of Glutaraldehyde: 100g/mol and Sodium Cacodylate:160g/mol

Assuming the density is 1 of Glutaldehyde, then 25%(1000g)=250g
250g/molecular weight(100g/mol)= 2.5mol/L-->2.5M

After this I have no clue which way to go. Any help would be great.
 
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Use M1V1=M2V2. Don't need to go back to moles, percentages are fine.
 
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