Asked by Jimmy K
You are asked to prepare 0.15 M HCl from a 36% W/v stock solution.Describe the procedure of the preparation.
Answers
Answered by
DrBob222
0.15 M HCl = 0.15 x 36.5 molar mass = 5.475 g HCl/L solution.
We have 36% w/v which means 36 g HCl/100 mL solution. Therefore, each mL of this solution contains 0.36 g HCl. How many mL must we take to contain 5.475 g HCl? We must take 5.475 g x (1 mL/0.36 g) = 15.21 mL of the 36% soln. Therefore, 15.21 mL of 36% HCl soln made to a final volume of 1 L will be 0.15 M HCl. Check.
15.21 mL x 36 g HCl/100 mL = 5.475 g HCl.
5.475 g HCl x (1 mol HCl/36.5 g HCl) = 0.15 M.
You should redo the problem but use the exact molar mass HCl. I rounded.
Check my thinking.
We have 36% w/v which means 36 g HCl/100 mL solution. Therefore, each mL of this solution contains 0.36 g HCl. How many mL must we take to contain 5.475 g HCl? We must take 5.475 g x (1 mL/0.36 g) = 15.21 mL of the 36% soln. Therefore, 15.21 mL of 36% HCl soln made to a final volume of 1 L will be 0.15 M HCl. Check.
15.21 mL x 36 g HCl/100 mL = 5.475 g HCl.
5.475 g HCl x (1 mol HCl/36.5 g HCl) = 0.15 M.
You should redo the problem but use the exact molar mass HCl. I rounded.
Check my thinking.
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