← Zurück zur vorhergehenden Übersicht: Säure-Base_en
- This topic has 1 reply, 1 voice, and was last updated 5 days, 19 hours ago by .
- Posts
Describe the manufacture of the following solutions in the laboratory: the solvent is always distilled water.
Adopted the substances be as salts of which: and solving itself fully in water.- (a) 1 litre of 0.3 M NaOH solution
- (b) 1 litre of 0.4 M HCL solution
- (c) 25 ml 0.5 M NaOH solution
- d) 30 ml 0.6 M H2SO4
Solution: The abbreviation 'M' is a note on the concentration, so I mean, z.B. 0.3 M, that 0.3 mol the the following: substance dissolved in one liter of water, in other words: 0.3 mol/l
With c=n/V and n=m/M follows: c=m/M/V respectively m=c·V·M
Respect: here is well, you know, the 'M' the molar mass ! Depending on the situation , the 'M' stands for concentration or molar mass !Action to be taken to the manufacture the solution: the solids, z.B. NaOH is weighted, in a z.B. Glass of glass given and so much distilled water as desired. Note that the solid requires a small volume of its own, but this need not be taken into account.
- a) m(NaOH)=c(NaOH)·V·M(NaOH) = 0.3 mol/l · 1.0 l · 58.5 g/mol = 17.55 g
Thus, 17.55 g of NaOH is weighed and the glass is filled with distilled water to the mark of 1.0 litres. - b) m(HCl)=c(HCl)·V·M(HCl) = 0.4 mol/l· 1.0 l · 36.5 g/mol = 14.6 g
So you have to weigh 14.6 grams of HCl and fill the glass with distilled water until you mark 1.0 liters. - c) m(NaOH)=c(NaOH)·V·M(NaOH) = 0.5 mol/l · 0.025 l · 58.5 g/mol = 0.73 g
Weighing 0.73 g of NaOH and filling it with water until 25 ml is marked. - d) m(H2SO4) = c(H2SO4) · V · M(H2SO4) = 0.6 mol/l · 0.03 l · 98 g/mol = 1.764 g
Weigh 1.764 g H2SO4 and fill with this water until the marking of 30 ml.
- You must be logged in to reply to this topic.