 The world’s only live instant tutoring platform Show About UsBecome a TutorBlog Filo instant Ask button for chrome browser. Now connect to a tutor anywhere from the web Add to Chrome Home Class 12 Chemistry Gases & Solutions Solutions Question 6982 Attempted students Solutions(1) Osmotic pressure symbol= [Molarity] where = solution constant Molarity= and moles= From this equation molar mass can be determined. 165 Connect with 50,000+ tutors in less than 60 seconds 24x7 Ask a TutorPractice questions - Asked by Filo studentsQuestion 1A solution containing 1.8 g of a compound (empirical formula ) in 40 g of water is observed to freeze at C. The molecules formulea of the compound is ( of water =1.86kg K): View Solution Question 2Each question contains
STATEMENT-I(Assertion) and STATEMENT-2(Reason).the statement carefully and mark the correct answer accoring to the instrution given below: View Solution Question 3The freezing point of 4% aqueous solution of 'A' is equal to the freezing point 10% aqueous solution of 'B'. If the molecules mass of 'A'is 60, then the molecules mass of 'B' will be: View Solution A solution of 0.640 g of azulene in 100.0 g of benzene is . The boilingpoint of benzeneis , and is /molal What is the moleculer mass of azulene? View Solution Are you ready to take control of your learning?Download Filo and start learning with your favourite tutors right away! 1. Important terms related to solutions: (i) Solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits. (ii) Solute and Solvent. In a binary solution, the component present in smaller amount is called solute while the other present in larger amount is called solvent. If water is the solvent, solution is called aqueous solution. (iii) Concentration terms of solutions (a) MolarityM= Moles of solute Vol. of solution (in mL) ×1000 (b) Molaritym= Moles of solute Mass of solvent (in g) ×1000 (c) Mole fractionx= Moles of solute Moles of solute+Moles of solvent (d) Mole fraction of solute+mole fraction of solvent=1 2. Laws governing solutions: (i) Henry’s law: The mass of the gas dissolved in a given volume of the liquid at constant temperature is directly proportional to the pressure of the gas in equilibrium with the liquid or for a mixture of gases in equilibrium with a liquid, the partial pressure of the gas is directly proportional to the mole fraction of the gas in the solution m=k.p or p=KH.x KH is Henry’s law constant. (ii) Raoult’s law in its general form can be stated as, for any solution the partial vapour pressure of each volatile component in thesolution is directly proportional to its mole fraction. 3. Colligative properties (i) Relative lowering in vapour pressure: p1 0-p1p1 0=x2 and M2=w2M1w1×p1 0 -p1p1 0 (ii) Elevation in boiling point ΔTb =Kb m and M2=Kb×w2×1000 ΔTb×w1 (iii) Depression in freezing point ΔTf=Kfm and M2=Kf×w2×1000ΔTf ×w1 (iv) Osmotic pressure π=cRT or π=nVRT and M2=w2RTVπ (v) Van't Hoff factor, i=Normal molar massAbnormal molar mass= Observed colligative property Calcluated colligative property (a) Relative lowering in vapour pressure Δpp=i xsolute (b) Elevation in boiling point, ΔTb=i Kbm (c) Depression in freezing point, ΔTf=i Kfm (d) Osmotic pressure, π=i nVRT How is the molar mass of a non volatile solute determined by osmotic pressure?∴π=mW×V1×S. T⟶ From this equation molar mass can be determined.
How do you find the molar mass of a non volatile solute?Its molar mass is 40 g/mol.. The number of moles of non volatile solute =40w=0. 025w moles.. 114 g of octane (molar mass 114 g/mol) corresponds to 114114=1 mole.. The mole fraction of non volatile solute is =1+0. 025w0. 025w. ... . P0P0−P=XB.. 10020=1+0. 025 w0. 025w.. 100(0. 025w)=20(1+0. 025w). 5w=20+0. 50w.. 2w=20.. How do you calculate molar mass of non volatile solute using vapour pressure lowering?MB=wB×MA×PSwA×(P∘A−PS)
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Derive an expression to calculate molar mass of non volatile solute by osmotic pressure
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