How can 1 M of HCl be prepared from 32 percent HCl concentration?

Answer 1

percentage = well, mostly weight percentage is the most commonly given percentage density = the density of a sol'n is constant, for HCl, the value is 1.18 g/ mL now assuming you have 34% of HCl in the sol'n in preparing 1M HCl, calculate the mass of HCl Needed.

assuming there are 100g of the whole solution, therefore there will be 34g of HCl:

percent by weight = (34% x 100g) / 100g

= 34g

now to get the mass, we need dimensional analysis. 34 g HCl x 1 mol HCl x 1 L sol'n x 1.18 g HCl x 1 mL = FINAL ANSWER

36.45 g HCl 1 mol HCl 1 mL 1 L sol'n you just get the final answer.... percentage = well, mostly weight percentage is the most commonly given percentage density = the density of a sol'n is constant, for HCl, the value is 1.18 g/ mL now assuming you have 34% of HCl in the sol'n in preparing 1M HCl, calculate the mass of HCl Needed.

assuming there are 100g of the whole solution, therefore there will be 34g of HCl:

percent by weight = (34% x 100g) / 100g

= 34g

now to get the mass, we need dimensional analysis. 34 g HCl x 1 mol HCl x 1 L sol'n x 1.18 g HCl x 1 mL = FINAL ANSWER

36.45 g HCl 1 mol HCl 1 mL 1 L sol'n you just get the final answer....

Answer 2

You should also know the density of the 32 %(m/m) HCl.

Maybe found answer in 'Related questions':

"How do you prepare a_solution of 1 M HCL using 35 percent HCL", a link is just below this answer.

Answer 3

To prepare 1 M HCl solution from 32% HCl solution, you would need to dilute the 32% HCl solution with water. To calculate the dilution, you would use the formula C1V1 = C2V2, where C1 is the initial concentration (32%), V1 is the volume of the initial solution, C2 is the final concentration (1 M), and V2 is the final volume (1 liter for 1 M solution). By rearranging the equation and solving for V1, you can determine the volume of 32% HCl solution needed to prepare 1 M HCl solution.