how many grams H2SO4 required to prepare 1 litre of Semi molar solution?
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Explanation:
Mass of solute ‘w’ = 38 gram
Mass of Solution= 100 gram
Mass of Solvent ‘W’= 100-38
= 62 gm = 0.062 kg
m of HCl =1 +35.5 =36.5
molality = w/ mW(Kg.)
=38/(36.5 x 0.62)
molality = 16.79 mole /Kg.
Volume =mass /density=100/1.19 =84 ml. =0.084 liter
M = w/ mV(l)
=38 /(36.5 x 0.084)
Molarity = 12.39 mole/liter
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