Question

The Atomic Weight of a Metal Data Trial 2 Trial 1 .036a 37.3mL 22.8°C 33.5ml 22. 2°C 1. Mass of Magnesium (g) 2. Volume of Hydrogen Gas Collected (mL) 3. Temperature (°C) 4. Temperature (K) (show calculation) 5. Barometric Pressure (mm Hg) Height Difference between Water Levels (mm H,O) tin loin Mercury Equivalent of Height Difference Water Difference (show calculation) 13.6 Vapor Pressure of Water (will be given-mm Hg) Pressure of Dry Hydrogen Gas = Barometric Pressure - Mercury Equivalent of Height Difference - Vapor Pressure of Water (show calculation) 10. Number of Mole of Hydrogen Gas Collected 11. Number of Mole of Metal 12. Molar Mass of a Solid (g/mol) 13. Average Molar Mass 14. Percent of Error

Answer 1

Mass of mag) volume of 4) - Temporaturc ) ) 0.36 g 373mL. 22.8°C asrg 33.5mL 22.a'c 295.35 K 269.36 10 in Temperature in - 295.95k Kelvin Batomemc prosj4KXD 969.36 Height dilteance a 7 in water levels Mercury equiralent as a huge of Height difference(x) 13.6 0.514 10 :0.73589 3.6 13.6 19.03 Vapour pressure of — 21.07 H₂O (mm Hy) (2) Pressure of Dry - 769.36 -0.514- Hydrogen gas 21.07= 747.78 (x-x-2) No. of moles 769.36 -0.7353 – 19.03= 749.6

& Pressure of Dry Hydrogen gas (x-y-z) - Barometric pressure - mercury equivalent of mercung - rapor pressure of ho. - For Trial 1 - 769.36 - 0.514-21.07 >> 747.78 fro Trial a = 769-36-0.7353-19.03 > 949.5 Number of mole of Hydromen gas collected. - Density of Hydrogen. - 0.089889/2 - volume of He gas - 37.3 mL for trial & 33.8mL for trial 2

- mass= Volumex Donssly. - Trial = 37.3mL x 0.089889/L =) 37.3 x 10 Iye x0.0898891 3.352 x 103g Trial 2) 33.5 mL x 0.08988 g/L => 33.5x16?y* 0.08988 gly - 3.0109 1103g = molecular weight :- 1.007849 no. of moles for trial 1. - 3.352x10 & 3.3259x105 mol 1.0078 4 9 for trial a - 3.0109 x 10 g = 2.987x10 mol 1.009849

* Number of mole of metal, Inilah mass of magnesium at sal- - 0.036g molecular weight - 24.3059 i no of moles of metal.. 0.036g 24.305 ~ 1.481 m 10 mol - mass of magnesium -Ioiala a um 0.037g molecular weight - 24.305 g i no of moles of metal =) 0.0379 24.300g 2) 1.522x10 mol

* Mulay mass of solid (g/mol Trial 0.036g 1.481 x 103 mul → 24.30 o/mol a trial - 0.0379 1.522 x 10 mol → 24.31 g/mol just the 6 Average molar mass is mean of molar mass. → 24.30 + 34.31 24.305 g/mol.

& Percent emr Tral (24.305-24-30) x 100 0.5% Trial - (24.305 - 24.31 x 100 -0.5%