[eunChae]

I stuck into this problem in a general chem book:
A mixture of 1 g H2 and 10 g O2 is ignited in a 3.15 L flask. What is the pressure in the flask when it is cooled to 25 C?
My approach was the following:
1 g H2 is 0.5 mole and therefore 8 g 02 will be spent. There will be 2 g(0.0625 mole) O2(g) and 9 g H2O(l) at 25 Celsius.
After that, by the ideal gas equation, P*3,15 L = 0.0625 mol * 62,364 L*Torr*K^(-1) * 298 K ==> P=368,74 mmHg (the pressure of O2)
Also according to the table in my book, vapor pressure of water at 25 C is 23.8 mmHg. Totally, 368,74 + 23.8 =392.54 mmHg
However, in the answers part the answer is 4.0*10^2. 
So what am i doing wrong for Gods sake?
Please comment. Thanks in advance...
eunChae

[Borek]

This is the same value, just with different number of significant digits.

[eunChae]

ow, this came to my mind but i thought it should be 3.9*10^2 with two significant figures. I guess you re right. That is a relief. Thanks.