200. mL. of 0.250 M HCl from a 2.50 M HCl solution I would addmL of water to a v
ID: 580265 • Letter: 2
Question
200. mL. of 0.250 M HCl from a 2.50 M HCl solution I would addmL of water to a volumetric flask, then add mL of 2.50 M HCl to the flask, let it cool, then dilute with water to the -mL mark b S0.mL of 1.75 M H,So, from a 4.00 M HSO, solution I would addmLof water to a volumetric flask, then addmL of 4.00 M H,SO, to the flask with water, let it cool, then dilute to the -mL. mark 100. mL of normal saline solution, 0.69% (w/v) NaCl, from a 5.0% (w/v) NaCl solution I would add mL of 5.0% (w/v) NaCl to a volumetric flask, ten dilute to the -mL mark with water, I would be sure to shake it well. 025 mL of 7.00% (v/v) acetone from 21.0% (v/v) acetone I would add well. mL of 21.0% (v/v) acetone to a volumetric flask, then dilute to the .i--mL mark with water. I would be sure to shake itExplanation / Answer
a) For dilution
mmoles after dilution = mmoles of solute before dilution [you add only solvent]
Thus 200mL x 0.25 M = V mL x 2.5M
Thus the volume of 2.5M solution to be taken is 20 mL
Thus the fill in the blanks are 20mL , 20mL and 200mL respectively.
b) this is also dilution.
Thus
50mLx 1.75M = 4 x V mL
Thus volume of acid = 21.875 mL
The blanks are 10mL ,21.875mL and 50mL respectively.
c) xw/v % means 100mL of solution has x g of solute.
Thus 5% w/v NaCl means
100mmL solution has 5 g of NaCl solution.
We need to get 0.69% w/v solution. That is the mass of NaCl in 100 mL is 0.69 g
If 100mL of stock has 5g NaCl
what volume can have 0.69g
V = 100x0.69/5 = 13.8mL
Thus the gaps are 13.8mL and 100mL
d) %v/v is 21%
That is 21 mL of acetone is in 100mL of solution.
To have 7% solution
i need to take = 7x100/21 = 33.33mL of 21% solutio and dilute to 100mL.
Thus the blanks are 33.33mL and 125mL mark
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