Homework Solution: Question 5 (5 points) The normal freezing point of benzene (CgH6) is 5.48°C. The…

    Question 5 (5 points) The normal freezing point of benzene (CgH6) is 5.48°C. The normal freezing point of a solution prepared by dissolving 2.00 g of a covalent compound in 100.0 g of benzene is 5.08°C. what is the molar mass of the compound? For benzene, Kf = 5.12 °C/m. Hint: The covalent compound does not dissociate in benzene a) 130 g/mol b) 260 g/mol c) 2600 g/mol d) 26 g/mol 1240 AM 9/6/2017 ^脈
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    Question 5 (5 points) The normal freezing point of benzene (CgH6) is 5.48°C. The normal freezing point of a solution prepared by dissolving 2.00 g of a covalent compound in 100.0 g of benzene is 5.08°C. what is the molar mass of the compound? For benzene, Kf = 5.12 °C/m. Hint: The covalent compound does not dissociate in benzene a) 130 g/mol b) 260 g/mol c) 2600 g/mol d) 26 g/mol 1240 AM 9/6/2017 ^脈

    Expert Answer

    5) DeltaTf = Kf x m

    Question 5 (5 objects) The natural freezing object of benzene (CgH6) is 5.48°C. The natural freezing object of a discerption expert by dissolving 2.00 g of a covalent junction in 100.0 g of benzene is 5.08°C. what is the molar bulk of the junction? For benzene, Kf = 5.12 °C/m. Hint: The covalent junction does not attributable attributable attributable attributable attributable attributable withdraw in benzene a) 130 g/mol b) 260 g/mol c) 2600 g/mol d) 26 g/mol 1240 AM 9/6/2017 ^脈
    media%2Fbee%2Fbeebe0be-a1a7-4f06-8db7-dc

    Question 5 (5 objects) The natural freezing object of benzene (CgH6) is 5.48°C. The natural freezing object of a discerption expert by dissolving 2.00 g of a covalent junction in 100.0 g of benzene is 5.08°C. what is the molar bulk of the junction? For benzene, Kf = 5.12 °C/m. Hint: The covalent junction does not attributable attributable attributable attributable attributable attributable withdraw in benzene a) 130 g/mol b) 260 g/mol c) 2600 g/mol d) 26 g/mol 1240 AM 9/6/2017 ^脈

    Expert Tally

    5) DeltaTf = Kf x m

    DeltaTf = 5.48 – 5.08 = 0.4

    m = DeltaTf / Kf = 0.4 / 5.12 = 0.078

    molality = (W/MW) (1000 / bulk of solvent in g)

    0.078 = (2.0 / MW) (1000 / 100)

    2.0 / MW = 0.0078

    MW = 256.41 g /mol

    6) Pi = iCRT

    Pi = osmotic influence = ?

    i = 5 (Al2(SO4)3 gives 5 ions)

    C = 0.01 M

    R = 0.0821

    T = 298 K

    Pi = 5 x 0.01 x 0.0821 x 298

    Pi = 1.22 atm