Q1. 25 mL of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was
Solution
The concentration is expressed in parts per million (ppm) when one part of solute is dissolved in one million parts of solvent
The concentration is expressed in parts per million (ppm) when one part of solute is dissolved in one million parts of solvent
Q2.1.0 g of a non-electrolyte solute (molar mass 250 g mol^(-1)) was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is 5.12 K kg mol^(-1), the lowering in freezing point will be ::
Solution
∆T=(1000 ×K_f×w)/(m×W) =(1000 ×5.12×1)/(250 ×51.2) = 0.4 K
∆T=(1000 ×K_f×w)/(m×W) =(1000 ×5.12×1)/(250 ×51.2) = 0.4 K
Q3. The mole fraction of the solute in one modal aqueous solution is
Solution
w=1000 g(H_2 O);n=1 mol N=W/M=1000/18=55.55 X_solute=n/(n+N)+1/(1+55.55) =0.018
w=1000 g(H_2 O);n=1 mol N=W/M=1000/18=55.55 X_solute=n/(n+N)+1/(1+55.55) =0.018
Q4. The osmatic pressure of a 5% (wt./vol) solution of cane sugar at 150℃ is
Solution
C=5/342×1/100×1000=50/342 mol/L Ï€=50/342×0.082×423=5.07 atm
C=5/342×1/100×1000=50/342 mol/L Ï€=50/342×0.082×423=5.07 atm
Q5.Conc〖 H〗_2 SO_4 has a density of 1.98 g/mL and is 98% H_2 SO_4 by weight. Its normality is
Solution
Strength of H_2 SO_4=98×19.8 g/L S=eq.wt.×N N=S/(eq.wt.)=(98×19.8)/49=39.6
Strength of H_2 SO_4=98×19.8 g/L S=eq.wt.×N N=S/(eq.wt.)=(98×19.8)/49=39.6
Q6. After adding a solute freezing point of solution decreases to -0.186. Calculate ∆T_b if k_f=1.86 and k_b=0.512
Solution
Q7.The freezing point (in °C)of a solution containing 0.1 g of K_3 [Fe(CN)_6] (mol.wt.329) in 100 g of water is : (K_f=1.86 K kg mol^(-1))
Solution
Q8.A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. the vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be
Solution
Q9.The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is :
Solution
Q10. At 40 ͦC the vapour pressure in torr, of methyl alcohol-ethyl alcohol solutions is represented by the equation. P=119 X_A+135; where X_A is mole-fraction of methyl alcohol, then the value of lim┬(X_A⟶1)〖P_A/X_A 〗is :
Solution