Your 0.5 mol HCl needed and 18.25 g needed are correct but it's the long way of going about it. However, you have no connection between this work and the 37 g/100 that you have in the stock. You can complete your work by
mass = volume x density or
18.25 = volume x 1.19 or volume = 15.336 mL (and I know that's too many sig figures) of the stock BUT since that is only 37% pure, then
15.336/0.37 = about 41.4 mL of the stock you want to take.
The easier way to work this is to 1. Determine the molarity of the stock.
2. Use the dilution formula.
1.
M of the stock.
1.19 g/mL x 1000 mL x 0.37 = g HCl in 1000 mL. Then g/36.5 = about 12 M but you can be more accurate than that.
2. mL1 x M1 = mL2 x M2
500*1 = mL2 x 12.06M
mL2 = about 41 mL
Perpare 500ml of 1M HCl solution from a bottle of concentrated 37%(W/W) HCL stock with a density of 1.19g/ml.
37%= 37/100 x 100
therefore mass is 37g
molarity = mol/vol
1= mol/0.5
mol=0.5
mass of Hcl needed = molxmr
mass = 0.5 x 36.5
mass = 18.25g
1 answer
1 answer