Question

Use table 2.3 volume occupied by 1,000 g of heavy water in air against stainless steel weights to calculate the real volume (vreal) to 25 degrees Celsius

Chemicals, Apparatus, and Unit Operations of Analytical Chemistry TABLE 2-3 Volume Occupied by 1.000 g of Water Weighed in Air against Stainless Steel Weights* Volume, mL Temperature, T, °C Corrected to 20°C 1.0013 1.0014 1.0015 1.0016 1.0018 1.0019 1.0021 1.0022 1.0024 1.0026 1.0028 1.0030 1.0033 1.0035 1.0037 1.0040 1.0043 1.0045 1.0048 1.0051 1.0054 1.0016 1.0016 1.0017 1.0018 1.0019 1.0020 1.0022 1.0023 1.0025 1.0026 1.0028 1.0030 1.0032 1.0034 1.0036 1.0037 1.0041 1.0043 1.0046 1.0048 1.0052 12 13 16 17 18 19 20 21 23 24 25 26 27 28 29 30 Corrections for buoyancy (stainless steel weights) and change in container volume have been applied.

Answer 1

First we have to calculate 1000 g of heavy water is equivalent to how many grams of water.

At 25^oC...1 mole of H2O is equivalent to 1 mole of D2O

So x/18=1000/20

x = 900

1g of water at 25^oC has a volume of 1.0037 ml

therefore, 900 g of water at 25^oC will have a volume (V real) of 1.003*900 = 903.33 ml against stainless steel.