1) Given that mass of anthracene = 2.88 g molar mass of anthracene = 178.2 g/mol

Question

1)

Given that mass of anthracene = 2.88 g

molar mass of anthracene = 178.2 g/mol

mass of solvent ( biphenyl) = 25.74 g = 0.02574 kg

Then,

molality = ( mass of solute / molar mass of solute) x (1/ mass of solvent in Kg)

=  ( mass of anthracene / molar mass of anthracene) x (1/ mass of biphenyl in Kg)

= ( 2.88 g/ 178.2 g/mol) x (1/0.02574 Kg)

= 0.63 m

Therefore,

molality of the solution = 0.63 m

--------------------------------------------------------------------------------------

depression in freezing point Tf = 71.0 oC - 21.1 oC = 49.9 oC

We know that

   depression in freezing point Tf = Kf x molality

  Kf = Tf / molality

= 49.9 oC/ 0.63 m

= 79.2 oC/m

Therefore,

molal freezing point depression constant Kf =  79.2 oC/m

Explanation / Answer

Part ILLong Questions) Show all your calculations clearly 4 questions (30 points) 1. (6pt) Pure biphenyl Cm mass 154.2 g/mol freezes at 71.0°C. A solution of 2.88 g of anthraene (C14Ho molar mass 178.2 g/mol) and 25.74g of biphenyl froze at 21.1. Calculate the molality of the solution (m) and the molal freezing point depression constant (K. Show all calculations. Note: Kr AT/ molality 2 (8pt) A student synthesized 3.214g of Ni(NHCI, from 4.023g of NiC12 6H,O. He dissolved 0.81g of Ni(NH,),Cl, in 40 mL of 1.0M HC1 to make a sample for analysis. Molar mass: Ni 58.7; Cl, 71 NH, 17; HO 18 (1) A 0.10M Ni2+ standard solution had a %T of 36.5, while his sample had a %T of 41.6. Calculate the N concentration in his sample and the mass percent of Ni in the product. Note: A-2-log%T. (2) The student used 15.75 mL of 0.498M NaOH to neutralize excess HCl in 15mL of the sample. Calculate the mass percent of NH3 in the product.

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