DATA Trial a. Mass of Mg (g) b. Temperature of water 23 2 a 8 K 2 99 c. Temperat

Question

DATA Trial a. Mass of Mg (g) b. Temperature of water 23 2 a 8 K 2 99 c. Temperature of water (in Kelvins) Tab d. Volume of H2 gas generated (mL) in lab 2 m l conditions lab atmospheric pressure (from the electronic barometer) 2 172 in Hg lab(total) how the conversion from inches Hg to mmHg using 1 in.Hg 25.4 mmHg below: mm Hg alculate Pabaar) for each trial: P lab(total) lab(H20) lab(H2) See the example on the page 76, and assume Phbototary is the same for all three trials) Trial 1 2 3 lab H20) see web link on mmHg. mmHg mmHg 76 mmHg Piabau Vlab Te culate the STP volume of H2 using v. STr Refer to entries c, d, and lab PST om your Data Tables above and use the temperature and pressure at STP for TSTP and P ow your work for one trial below:

Explanation / Answer

From the data

Trial 1

T(lab) = 23 + 273 = 296 K

P(lab) = (29.77 x 25.4 - 21.1)

          = 735.06 mmHg/760 mmHg

          = 0.97 atm

V(lab) = 0.012 L

We get,

V(stp) = 0.97 x 0.012 x 273.15/296 x 1 = 0.0107 L

moles of H2 = 0.0371/24.305 = 0.00153 mol

molar volume = 0.0107/0.00153 = 7.01 L/mol

Trial 2

T(lab) = 24.5 + 273 = 297.5 K

P(lab) = (29.77 x 25.4 - 23.32)

          = 732.84 mmHg/760 mmHg

          = 0.96 atm

V(lab) = 0.016 L

We get,

V(stp) = 0.96 x 0.016 x 273.15/297.5 x 1 = 0.0141 L

moles of H2 = 0.0349/24.305 = 0.00144 mol

molar volume = 0.0141/0.00144 = 9.792 L/mol

Average molar voluem of H2 = 8.401 L/mol

% error with 22.4 L/mol as true value = (22.4-8.401) x 100/22.4 = 62.49%

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