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Chapter 10: Problem 20

You need 70.2 \(\mathrm{J}\) to raise the temperature of 34.0 \(\mathrm{g}\) of ammonia, \(\mathrm{NH}_{3}(g),\) from \(23.0^{\circ} \mathrm{C}\) to \(24.0^{\circ} \mathrm{C}\) . Calculate the molar heat capacity of ammonia.

Short Answer

Expert verified
The molar heat capacity of ammonia is \(35.1 \, J/mol \cdot ^{\circ}C).

Step by step solution

01

Identify all given values

The heat transferred (\(q\)) is 70.2 J. The initial temperature is \(23.0^{\circ} C\) and the final temperature is \(24.0^{\circ} C\), so the change in temperature (\(\Delta T\)) is \(24.0 - 23.0 = 1.0 \(^{\circ}C\). The mass of ammonia (\(NH_{3}\)) is 34.0 g.
02

Calculate the amount in moles

To compute for the number of moles (\(n\)), one applies the formula \(n=\frac{m}{M}\) where \(m\) is the mass and \(M\) is the molar mass. The molar mass of ammonia (\(NH_{3}\)) is 17.03 g/mol. So, \(n=\frac{34.0 \, g}{17.03 \,g/mol} = 2.0 \, mol \).
03

Utilize the molar heat capacity formula

Substitute the values calculated in the formula \(C = \frac{q}{\Delta T \times n} \), obtaining \(C = \frac{70.2 \, J}{1.0 \, ^{\circ}C \times 2.0 \, mol} = 35.1 \, J/mol \cdot ^{\circ}C). This value represents the molar heat capacity of ammonia.

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