the vapour pressure of dilute aqueous solution of glucose is 750 mm of HG at 373 Kelvin vapour pressure of pure solvent is 760 mm of HG calculate molality and mole fraction of solute
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Molality, M = moles of solute/kg of solvent
Mole fraction, xA = nA/nA+nB, xB= nB/nA+nB
p1 = x1po1
∴ x1 = p1/po1 = 750/760 = 0.9868
x2(solute) = 1 – 0.9868 = 0.0132
molality, m = x2/x1M1 * 1000 = 0.0132 * 1000/0.9868 * 18 = 0.7503 mol kg-1
ALTERNATIVESOLUTION :
Given that :
Temperature = 273 K
boiling point of H2O = 373 K
∴ vapour pressure H2O = 76 cm
We have,
Po-Ps/Ps = w*M/w*M
∴ molality
= w/w*M * 1000 = Po-Ps/Ps * 1/M * 1000
= 760 -750/750 * 1/18 * 1000
= 0.741 mol/kg of solvent
Also we have,
= 1000 *1.72 *20/50 *2
= 344
van’t Hoff factor (i) = actual mol. wt./calculate mol. wt = 172/344 = 0.5
Po-Ps/Ps = n/n +N
∴ mole fraction = Po-Ps/Po = 760 – 750/760
= 10/760 = 0.013
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Which of the folloPlease find the answer to your question
Molality, M = moles of solute/kg of solvent
Mole fraction, xA = nA/nA+nB, xB= nB/nA+nB
p1 = x1po1
∴ x1 = p1/po1 = 750/760 = 0.9868
x2(solute) = 1 – 0.9868 = 0.0132
molality, m = x2/x1M1 * 1000 = 0.0132 * 1000/0.9868 * 18 = 0.7503 mol kg-1
ALTERNATIVESOLUTION :
Given that :
Temperature = 273 K
boiling point of H2O = 373 K
∴ vapour pressure H2O = 76 cm
We have,
Po-Ps/Ps = w*M/w*M
∴ molality
= w/w*M * 1000 = Po-Ps/Ps * 1/M * 1000
= 760 -750/750 * 1/18 * 1000
= 0.741 mol/kg of solvent
Also we have,
= 1000 *1.72 *20/50 *2
= 344
van’t Hoff factor (i) = actual mol. wt./calculate mol. wt = 172/344 = 0.5
Po-Ps/Ps = n/n +N
∴ mole fraction = Po-Ps/Po = 760 – 750/760
= 10/760 = 0.013
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The vapor pressure of a solution is equal to the vapor pressure of the pure solvent times the mole fraction of solvent in the solution. So, since 373 is the normal boiling point of pure water, its vapor pressure at 373 is 760 mm Hg.
Psoln = PH2O XH2O
750 = 760 (XH2O)
XH2O = 0.987
The mole fraction of water is equal to moles water / (moles water + moles glucose). Because 1.00 kg H2O = 55.6 moles H2O,
0.987 = 55.6 /(55.6 + x)
x = moles glucose = 0.777 moles
Now, since molality = moles solute / kg solvent, molality = 0.777 molal