The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol ∙ K. The conversion of 50.0 g of ice at 0.00 °C to liquid water at 0.00°C requires ________ kJ of heat.

Question

Quizplus · Accepted Answer

To calculate the heat required to convert 50.0 g of ice to liquid water, we need to use the following formula:

q = m × ΔHfus

where

q = heat required (in kJ) 
m = mass of substance (in g) 
ΔHfus = heat of fusion (in kJ/mol)

First, we need to determine the number of moles of water that are present in 50.0 g of ice:

n = m ÷ M

where

n = number of moles 
m = mass (in g) 
M = molar mass (in g/mol)

The molar mass of water is 18.02 g/mol, so:

n = 50.0 g ÷ 18.02 g/mol = 2.78 mol

Now, we can use the formula to calculate the heat required:

q = 2.78 mol × 6.01 kJ/mol = 16.7 kJ

Therefore, the answer is B (16.7).

The Heat of Fusion of Water Is 6