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bazaltina [42]
1 year ago
13

How many moles of solute would be dissolved in .500 kg of solvent to make a 2.50 molal NaOH solution?

Chemistry
1 answer:
Tom [10]1 year ago
3 0

Molality is a measure of concentration that relates the moles of solute to the kilograms of solvent, it is described by the following equation:

Molality=\frac{molSolute}{kgSolvent}

We are given the molality(2.50m) and kilograms of solvent(0.500kg), so we solve for moles of solute from the equation:

molSolute=Molality\times kgSolvent\begin{gathered} molSolute=2.50m\times0.500kg \\ molSolute=1.25molSolute \end{gathered}

To make a 2.50molal NaOH solution would be needed 1.25moles of solute

Answer: 1.25 moles

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Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}     ------>NO_{(g)}

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

where:

\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

= 1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

\delta H^0__{R,435K} ≅ 91383 J

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