Question

To vaporize 2.00 g of ammonia 656 calories are required. How many kilojoules are required to vaporize the same mass of ammonia?

Answer 1

To convert calories into kilojoules, we can use the conversion factor 1 calorie = 0.004184 kilojoules.

First, we need to calculate the amount of heat required to vaporize 2.00 g of ammonia in kilocalories. We can use the formula:

Heat (calories) = Mass (g) * Heat of Vaporization (cal/g)

Given:
Mass = 2.00 g
Heat (calories) = 656 calories

Heat in kilocalories = Heat (calories) / 1000
Heat in kilocalories = 656 calories / 1000 = 0.656 kilocalories

Next, we convert the heat in kilocalories to kilojoules using the conversion factor:

Heat (kilojoules) = Heat (kilocalories) * 4.184
Heat (kilojoules) = 0.656 kilocalories * 4.184 = 2.738 kilojoules

Therefore, approximately 2.738 kilojoules are required to vaporize the same mass of ammonia.