I am having issue with letter b, but the context from letter a is needed to solve. I have provided my answer to a and the steps how I got there.   a. Calculate the density of H2O vapor at 1.00 atm and 100.0 °C in g/L. Circle your answer. My work: 100.0 °C + 273 K = 373 K d = (18.01528 g/mol × 1.00 atm)/( 0.08206 L atm mol-1 K-1 × 373 K) d = 18.01528 / 30.60838 d = 0. 589 g/L   b. 1.00 mL of water at 25 °C is heated to 100 °C, at which point it boils at an atmospheric pressure of 1 atm and is vaporized.  What is the difference in volume (in mL) when this happens?  (At 25 °C, liquid water has a density of 0.997 g/mL.) Circle your answer.

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I am having issue with letter b, but the context from letter a is needed to solve. I have provided my answer to a and the steps how I got there.

 

a. Calculate the density of H2O vapor at 1.00 atm and 100.0 °C in g/L. Circle your answer.

My work:

100.0 °C + 273 K = 373 K

d = (18.01528 g/mol × 1.00 atm)/( 0.08206 L atm mol-1 K-1 × 373 K)

d = 18.01528 / 30.60838

d = 0. 589 g/L

 

b. 1.00 mL of water at 25 °C is heated to 100 °C, at which point it boils at an atmospheric pressure of 1 atm and is vaporized.  What is the difference in volume (in mL) when this happens?  (At 25 °C, liquid water has a density of 0.997 g/mL.) Circle your answer.

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