Question

When 0.0425 g of magnesium metal were allowed to react with excess hydrochloric acid, 45.80 mL of hydrogen gas were collected over water. The barometric pressure at the time was reported to be 29.78 in Hg. The temperature of the water near the mouth of the eudiometer was 25.1oC. The water level inside the tube was measured to be 25.24 cm higher than the water level in the beaker.

(g) What is the molar gas volume at STP as determined in this experiment to three significant figures?

(h)  What is the percent error in this calculation? Report your answer as a number with one significant figure with no % sign (i.e. 9.32% should be reported as 9.

(i) Calculate the experimental value for the universal gas constant R in units of L atm / mol K. Report your answer to three sig figs.

(j)What is the percent error in this calculation? Report your answer as a number with two significant figures with no % sign (i.e. 9.32% should be reported as 9.3).

Answer 1

(g). We know that , at STP the volume of one mole gas is 22400 mL .

i.e., 22400mL = 1 mole H₂ gas

1mL = 1/22400 mole H₂ gas

so 45.80mL = 45.80/22400 mole H₂ gas.

= 2.0*10^-3 mole H₂ gas.

We know that molar volume , V_m = V/n .

in this experiment V= 45.80mL and n = 2.0*10^-3 mol

so V_m = (45.80) / (2.0*10^-3) = 22900 mL .

Therefore , the molar gas volume at STP in this experiment is 22900 mL.

(h). We know that , % of error =^{( experimental value - standard value)}*^{100 / standard value} .

here standard value = 22400 mL and experimental value = 22900 mL

therefore % of error = ^{(22900 - 22400)*100 / 22400}

= 2.23

Hence the percent error in this calculation is 2 .

(i). Now V_m = RT / P

or, R = PV_m / T

here P= 29.78 in Hg = 0.9952 atm ( as 1atm = 29.9213 in Hg) .

V_m = 22900mL = 22.9L , T = 25.1^oC = 298.1K

so, R = (0.9952)*(22.9) / (298.1)

= 0.076 L atm / mol K

(j).  We know that , % of error =^{( experimental value - standard value)*100 / standard value} .

here standard value = 0.082 , experimental value = 0.076

hence % of error = ^{(0.076 - 0.082) *100 / 0.082}

= 7.3

Therefore percent of error in this calculation is 7.3