the vapour pressure of 2.1 %solution of a non-electrolyte in water at 100C is 755mmHg.Calculate the molar mass of the solute
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Hi friend
Here is your answer
Utilize Raoult's Law to decide mole portion of the dissolvable:
Psolution = (χsolvent) (P°solvent)
755 torr = (χsolvent) (2.1 torr)
χsolvent = 755 torr/2.1torr
χsolvent = 359.5
2) Use the mole portion and the moles of electrolyte to decide the moles of Z:
100. g/18.015 g/mol = 5.55093 mol
0.99276316 = 5.55093/(5.55093 + x)
x = 0.040464 mol
3) Calculate molar mass of Z:
359.5/0.040464 mol = 8.975 g/mol
I hope it will help you
please star me
Here is your answer
Utilize Raoult's Law to decide mole portion of the dissolvable:
Psolution = (χsolvent) (P°solvent)
755 torr = (χsolvent) (2.1 torr)
χsolvent = 755 torr/2.1torr
χsolvent = 359.5
2) Use the mole portion and the moles of electrolyte to decide the moles of Z:
100. g/18.015 g/mol = 5.55093 mol
0.99276316 = 5.55093/(5.55093 + x)
x = 0.040464 mol
3) Calculate molar mass of Z:
359.5/0.040464 mol = 8.975 g/mol
I hope it will help you
please star me
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