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QUESTION

# A sample of ammonia liberates 5.66 kJ of heat as it solidifies at its melting point. What is the mass of the sample? DeltaH_(solid)  = -5.66 kJ/mol?

It's important to realize that the "5.66 kJ" of heat released is actually q, the heat flow, not the of freezing, DeltaH_"frz".

"Solidification at the melting point" is otherwise known as freezing at the freezing point, so this process occurs at T_f = -77.73^@ "C" for ammonia.

The equation that relates enthalpy to heat flow at constant pressure is:

\mathbf(DeltabarH_"frz" = q_p/n_"compound")

where:

• DeltabarH_"frz" is the molar enthalpy of freezing, in "kJ/mol".
• q_p is the heat flow q at a constant pressure.
• n_"compound" is the "mol"s of the compound that is freezing.

So, all you need to realize is that you know q_p and DeltaH_"frz". Therefore:

n_"compound" = q_p/(DeltaH_"frz")

= (-"5.66 kJ")/(-"5.66 kJ/mol")

= color(blue)("1 mol")

Thus, the mass of ammonia is:

= "1 mol" xx (14.007 + 3xx1.007"9 g NH"_3)/("1 mol NH"_3)

= color(blue)("17.0307 g NH"_3)