How many milliliters of 1.9 M HCl must be

transfered from a reagent bottle to provide
24 g HCl for a reaction?

To calculate the volume of 1.9 M HCl needed to provide 24 g of HCl, we can use the formula:

Volume (in L) = Number of moles / Molarity

First, let's determine the number of moles of HCl required for the reaction using its molar mass. The molar mass of HCl is approximately 36.46 g/mol.

Number of moles = Mass / Molar mass
Number of moles = 24 g / 36.46 g/mol ≈ 0.658 mol

Next, we can use the formula mentioned earlier to calculate the volume:

Volume (in L) = Number of moles / Molarity
Volume (in L) = 0.658 mol / 1.9 mol/L ≈ 0.346 L

Since the question asks for the volume in milliliters, we need to convert L to mL. There are 1000 mL in 1 L.

Volume (in mL) = Volume (in L) * 1000
Volume (in mL) = 0.346 L * 1000 ≈ 346 mL

Therefore, approximately 346 milliliters of 1.9 M HCl must be transferred from the reagent bottle to provide 24 g of HCl for the reaction.

volume*massHCL/molmass =1.9moles/liter

volume= 1.9*molmassHCl/massneeded