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CaHeK987 [17]
3 years ago
9

Quinine is a natural product extracted from the bark of the cinchona tree, which is native to South America. Quinine is used as

an antimalarial agent. When 3.28 g of quinine is dissolved in 25.0 g of cyclohexane, the freezing point of the solution is lowered by 8.43 °
c. The freezing point and Kf constant for cyclohexane can be found here.
Calculate the molar mass of quinine.
Chemistry
1 answer:
evablogger [386]3 years ago
5 0
For the answer to the question above, <span>The formula for freezing point depression is </span>
<span>ΔTf = mkfi </span>

<span>kf is the freezing point constant </span>
<span>i is the Van't Hoff factor which in this case is 1 </span>
<span>m is molality (moles of solute/kg of solvent) </span>
<span>ΔTf is temperature change </span>

<span>ΔTf is 2.17 °C, the molality is the amount of solute Quinine </span>
<span>in the solvent cyclohexane. We cannot calculate moles therefore we need to substitute moles with g/mm. </span>

<span>moles = g/mm so molality=(g/mm)/kg </span>
<span>molality = (0.845/mm)/0.025 = 33.8/mm </span>

<span>2.17 = 33.8/mm(20.8) rearrange </span>

<span>mm = (33.8/2.17)(20.8) = 324g/mol</span>
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Answer:

<u>The standard enthalpy of reaction = -4854.7kJ</u>

<u>The difference: </u>ΔH-ΔE = Δ(PV) = Δn.R.T = <u>9910.288 J ≈ 9.91 kJ</u>    

Explanation:

<u>The balanced chemical equation for the combustion of heptane</u>:

C₇H₁₆ (l) + 11 O₂ (g) → 7 CO₂ (g) + 8 H₂O (l)

Given: The standard enthalpy of formation (\Delta H _{f}^{\circ }) for: C₇H₁₆ (l) = -187.8 kJ/mol, O₂ (g) = 0 kJ/mol, CO₂ (g) = -393.5 kJ/mol, H₂O (l) = -286 kJ/mol

<u>To calculate the standard enthalpy of reaction (\Delta H _{r}^{\circ }) can be calculated by the Hess's law</u>:

\Delta H _{r}^{\circ } = \left [\sum \nu \cdot\Delta H _{f}^{\circ }(products)  \right ] - \left [\sum \nu\cdot\Delta H _{f}^{\circ }(reactants)  \right ]

Here, \nu is the stoichiometric coefficient

⇒ \Delta H _{r}^{\circ } =

\left [ 7\times \Delta H _{f}^{\circ }\left (CO_{2}\right )+ 8\times \Delta H _{f}^{\circ }\left (H_{2}O \right )\right ]

- \left [1\times \Delta H _{f}^{\circ }\left (C_{7}H_{16}\right ) +11\times \Delta H _{f}^{\circ }\left (O_{2} \right ) \right ]

=\left [ 7\times \left (-393.5 kJ/mol \right )+ 8\times \left (-286 kJ/mol \right )\right ]

-\left [1\times \left (-187.8 kJ/mol \right ) +11\times \left (0 kJ/mol \right ) \right ]

⇒ \Delta H _{r}^{\circ } = \left [ \left (-2754.5 \right )+ \left (-2288 \right )\right ]\left -[ \left (-187.8 \right ) +\left (0 \right )\right ]

⇒ \Delta H _{r}^{\circ } = \left [ -5042.5 ]\left -[ -187.8] = \left ( -4854.7kJ \right )

<u>To calculate the difference: </u>ΔH-ΔE=Δ(PV)

We use the ideal gas equation: P.V = n.R.T

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T

Given: Temperature:T = 298K, R = 8.314 J⋅K⁻¹⋅mol⁻¹

Δn = number of moles of gaseous products - number of moles of gaseous reactants = (7)- (11) = (-4)

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T = (-4 mol) × (8.314 J⋅K⁻¹⋅mol⁻¹) × (298K) = <u>9910.288 J = 9.91 kJ</u>                              (∵ 1 kJ = 1000J )

                                                                             

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