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Describe the manufacture of the following solutions in the laboratory: the solvent is always distilled water.
Adopted The substances are present as salt and dissolve completely in water.- a) 1 1 litre 0.3 M NaOH-Solution
- b) 1 1 litre 0.4 M HCL-Solution
- c) 25 Milligrams 0.5 M NaOH-Solution
- d) 30 Milligrams 0.6 M H2SO4
Lösung: The abbreviation 'M' is an indication of the concentration, for example. 0.3 M, that 0.3 mol the following substance dissolved in one litre 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
Attention, here's the thing 'M' Depending on the situation , 'M' Concentration or the mud !Process to produce the solution: the solid, e.g. NaOH Weighing, for example, in a cup glass and so much distilled water as desired. Note that the solid requires a small volume of its own, but this does not It must 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
So they will be 17.55 g NaOH Weighing and distilling the glass with water until the marking of 1.0 Liters filled. - b) m(HCl)=c(HCl)·V·M(HCl) = 0.4 mol/l· 1.0 l · 36.5 g/mol = 14.6 g
You have to 14.6 Grams HCl Weighing and distilling the glass with distilled water until the marking of 1.0 Fill up the liters. - c) m(NaOH)=c(NaOH)·V·M(NaOH) = 0.5 mol/l · 0.025 l · 58.5 g/mol = 0.73 g
weighting of 0.73 g NaOH and with that water until the marking of 25 fill in ml. - d) m(H2SO4) = c(H2SO4) · V · M(H2SO4) = 0.6 mol/l · 0.03 l · 98 g/mol = 1.764 g
weighting of 1.764 g H2SO4 and with that water until the marking of 30 fill in ml.
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