calculate If you know K_b 1.61 Times 10^-5 for cyanide, CN^-, you can calculate

Question

calculate If you know K_b 1.61 Times 10^-5 for cyanide, CN^-, you can calculate the equilibrium constant, K_a, for this reaction HCN + H_2O rightarow CN^- + H_3O^+ Lysine is an amino acid that is an essential part of nutrition but which is not synthesized by the human body. What is the molar mass of lysine if 750.0 ml. of a solution containing 8.60 g of lysine has an osmotic pressure of 1.918 atm? Calculate the temperature of an argon sample at 55.4 k Pa and 18.6 L if it occupies 25.8 L at 75.0 degree C and 41.1 kPa. A 3.05-g sample of an alloy of gold (Au) and copper is reacted with excess nitric acid to form 4.00 g of copper (II) nitrate. Thus the percentage of Au in the alloy, by mass, is The pH of 1.0M aqueous benzoic acid is 2.09. What is the value of K_a for benzoic acid? You have a 10.40-g mixture of table sugar (C_12H_22O_11) and table salt (NaCl). When this mixture is dissolved in 150. g of water, the freezing point is found to be -2.24 degree C. Calculate the percent by mass of sugar in the original mixture.

Explanation / Answer

31) Ka = Kw/Kb = (1*10-14)/(1.61*10-5) = 6.211*10-10

Thus, the correct option is : - (b)

32) Osmotic Pressure,P = C*R*T

or, 1.918 = C*0.0821*298

or, Molar concentration,C = 0.0784

Now, moles of lysine = molar concentration*volume of solution in litres = 0.0784*0.75 = 0.059

Now, molar mass of lysine = mass/moles of lysine = 8.6/0.059 = 146.268 g/mole

hence the correct option is :- (b)

33) Since the moles of Argon sample remains constant

Therefore, applying Combined gas Law i.e. (P1*V1)/T1 = (P2*V2)/T2

or, (55.4*18.6)/T1 = (41.1*25.8)/348

or, T1 = 338.174 K = 65.174 0C

34) Molar mass of Cu(NO3)2 = 187.5 g/mole

Thus, moles of Cu(NO3)2 in 4 g of it = mass/molar mass = 4/187.5 = 0.0213

Thus,moles of Cu in Cu(NO3)2 = Moles of Cu(NO3)2 = 0.0213

Thus, mass of Cu in the sample = moles*molar mass = 0.0213*63.5 = 1.355g

thus, % of Cu in the alloy = (mass of Cu/mass of alloy)*100 = (1.355/3.05)*100 = 44.415 %

Thus, % of Au in the alloy = 100 - 44.415 = 55.585%

thus, the correct option is :- (d)

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