at 373k the aqeous solution of glucose having vapour pressure of 750mmhg then calculate the mole fraction of the solute and molality of solution
Answers
Answered by
2
mark me as brainliest. please I am going to next rank
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
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
Similar questions