The molality of a 1L solution with x% H2SO4 is 9. Weight of the solvent present in sol. is 910g.Value of x?
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Best Answer: The molality of a solution is 9 when there are 9 moles by kg of solvent. For 910g =0.91kg there are 0.91*9= 8.19 moles of H2SO4 by kg of water. The molar weight of H2SO4=2*1+32+4*16=98g and so you have 98*8.19=802.62g =0.80262 of H2SO4 for 1 kg of water
the proportion in weight of H2SO4 is p= 0.80262/1+0.80262=0.445
so if you have a solution, with a density of 1g/cc, this corresponds to1000g in a litre or 1kg
x=445/1000=0.445=44.5%
Remark important. such a high concentration has a density much higher than 1g/cc!!!
the proportion in weight of H2SO4 is p= 0.80262/1+0.80262=0.445
so if you have a solution, with a density of 1g/cc, this corresponds to1000g in a litre or 1kg
x=445/1000=0.445=44.5%
Remark important. such a high concentration has a density much higher than 1g/cc!!!
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