Determine the amount of Sodium Salicylate required to make 100mL of a 0.01M stock solution. {this is from the spectrometric analysis of a commercial aspirin tablet lab}
c = 0.01 mol/L; v = 0.10 L; n = cv
∴n = 0.01*0.10 = 0.001 mols
MM = 160.11 g/mol; n = 0.001 mols; m = MMn
∴m = 160.11*0.001 = 0.16011 g
∴0.16011g of Sodium Salicylate is required to make 100 mL of a 0.01M Stock solution.
Helped me remember what we did in gr 11 I can't thank you enough man
thank you
To determine the amount of Sodium Salicylate required to make a 100 mL of a 0.01M stock solution, you need to use the formula:
Amount (in moles) = Concentration (in moles per liter) × Volume (in liters)
First, let's convert the volume to liters:
100 mL = 100/1000 = 0.1 liters
Now, plug in the values into the formula:
Amount (in moles) = 0.01 M × 0.1 L
Calculating this, we find:
Amount (in moles) = 0.001 moles
The molar mass of Sodium Salicylate (NaC7H5O3) is approximately 160.11 g/mol. To find the amount in grams, we use the formula:
Amount (in grams) = Amount (in moles) × Molar Mass
Plugging in the values:
Amount (in grams) = 0.001 moles × 160.11 g/mol
Calculating this, we find:
Amount (in grams) = 0.16011 grams
Therefore, you will need approximately 0.16011 grams of Sodium Salicylate to make 100 mL of a 0.01M stock solution for the spectrometric analysis of a commercial aspirin tablet lab.