Question

Yor u UELOW ! 35. What is the molar mass of a nonelectrolyte if 6.02 grams dissolved in 30.0 grams of benzene no va saven mol Sur freezes at -1.55 °C? The freezing point of pure benzene is 5.50 °C and the freezing point depression constant, Krp, is -5.12 Cm. ATEMki mm? a) 146 g/mol b) 201 g/mol c) 261 g/mol d ) 276 g/mol e) 376.2 g/mol Tsol- 5.50 = (-5.12) (X) Answer: a 0.03Kg 26 What is the heilinn enint of a lutinn nantsinine 122 of affeine C.H.N.Os dissolved in

Answer 1

formula depre ss ian 35 point we know, freezina non-electrolyte solute : k Cm Cm molal concentration a Forr ATT (Fina- Te(initial) T Cinitial) - 5.50 c Hesre, T(Final) 55 C 5.12 C/m ven Cm mo. OT moles Solute in 1000 am & the solvert 23 weiaht solute (in gm). W weight of solvert (in am) (W/M2) CWi/1000 solute (in gm/mole) mmolal M2 molon mass d N2 6.02 m N2/M2) T/1000) Thenefore, AT kEx WI 30 Caiven) 6.02/M2 30/1000 or, -55-5s0C 5-12 Cmx 6-02 x 100o -F.05 C -5.12 °C/m M2 x 30

6.02 xICOTO gm/mole 5 12 M2 7.05 30 M2 145.73 am/mole 146 m/mole oreslt Hence, mmolan mass M2=146 am/mole 29 calculatin vapor presswe trotl solutian mon electralyte solute (here suerose the with formula is P P NIM2 W2M Srii0 po aHi M2 P /po WIM2 the of mixcture P= vapor pressune pure solvert (here waten) pVapos the pfessune mm Ha (aiven) 42.2 15 am (aiven). solute weight weight of-solvent of 85 m (aiven) soluute 342 3 am/mole (given) M2 Molan mass Jolvert (t waten) 1 8 mars dt Molan MI- 15 x 18 Hemce 42.2 mm Ha x Po 85 x 342.3 41.&

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