Calculate the molar mass of nitrogen gas if 0.250 g of the gas occupies 46.65 ml at stp

pre-1982 definition STP: 120 g/mol post-1982 definition STP: 122 g/mol The answer to this question depends upon which definition of STP you're using. The definition changed in 1982 from 273.15 K at 1 atmosphere to 273.15 K at 10000 pascals. As a result the molar volume of a gas at STP changed from 22.4 L/mol to 22.7 L/mol. So let's calculate the answer using both definitions and see if your text book is 35 years obsolete. First, determine the number of moles of gas you have. Do this by dividing the volume you have by the molar volume. So pre-1982: 0.04665 / 22.4 = 0.002082589 mol post-1982: 0.04665 / 22.7 = 0.002055066 mol Now divide the mass you have by the number of moles. pre-1982: 0.250 g / 0.002082589 mol = 120.0428725 g/mol post-1982: 0.250 g / 0.002055066 mol = 121.6505895 g/mol Finally, round to 3 significant figures: pre-1982: 120 g/mol post-1982: 122 g/mol These figures are insanely large for nitrogen gas. So let's see if our input data is reasonable. Looking up the density of nitrogen gas at STP, I get a value of 1.251 grams per liter. The value of 0.250 grams in the problem would then imply a volume of about one fifth of a liter, or about 200 mL. That is over 4 times the volume given of 46.65 mL. So the verbiage in the question mentioning "nitrogen gas" is inaccurate at best. I see several possibilities. 1. The word "nitrogen" was pulled out of thin air and should be replaced with "an unknown" 2. The measurements given are incorrect and should be corrected. In any case, if #1 above is the correct reason, then you need to pick the answer based upon which definition of STP your textbook is using.

The answer is 4 i don't know if its right let me see.

Hope you find the answer