Aflexible container is put in a deep freeze. its original volume is 3.00 m3 at 25.0°c. after the container cools, it has shrunk to 2.00 m3. its new temperature in degrees celsius is

This is an ideal gas problem. The gas inside the balloon is considered ideal. Ideal gas equation is a function pressure, temperature, amount and volume. Note: amount is constant since the balloon ins closed. Pressure is maintained constant since the walls are flexible. Ideal gas equation is: PV=nRT. Put all constant in one side and variables in one. 

P/nR=T/V. To find the answer to the question equate the constants of both situation 
T1/V1=T2/V2
(25+273.15)/3=(x+273.15)/2
x=-74.38 degC 

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml