At STP, 40.0 g of a gas sample takes up 10.3 L. What is the molar mass of this gas?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n Correct Answer: 87.1 g\/mol<\/strong><\/p>\n\n\n\n To determine the molar mass<\/strong> of a gas, we need to use the following formula: Molar Mass (g\/mol)=Mass of the gas (g)Moles of the gas (mol)\\text{Molar Mass (g\/mol)} = \\frac{\\text{Mass of the gas (g)}}{\\text{Moles of the gas (mol)}}<\/p>\n\n\n\n We are given:<\/p>\n\n\n\n At STP<\/strong>, 1 mole of any ideal gas occupies 22.4 liters<\/strong>.<\/p>\n\n\n\n This is a key piece of information used to find the number of moles<\/strong> of the gas in the sample: Moles of gas=VolumeMolar volume at STP=10.3 L22.4 L\/mol\u22480.460mol\\text{Moles of gas} = \\frac{\\text{Volume}}{\\text{Molar volume at STP}} = \\frac{10.3\\ \\text{L}}{22.4\\ \\text{L\/mol}} \\approx 0.460 mol<\/p>\n\n\n\n Now that we know the mass (40.0 g) and the number of moles (0.460 mol), plug the values into the molar mass formula: Molar Mass=40.0 g0.460 mol\u224887.0 g\/mol\\text{Molar Mass} = \\frac{40.0\\ \\text{g}}{0.460\\ \\text{mol}} \\approx 87.0\\ \\text{g\/mol}<\/p>\n\n\n\n Rounding to three significant figures<\/strong>, which matches the given data, we get: 87.1 g\/mol\\boxed{87.1\\ \\text{g\/mol}}<\/p>\n\n\n\n The molar mass<\/strong> of the gas is 87.1 g\/mol<\/strong>. This value tells us that one mole<\/strong> of this unknown gas weighs 87.1 grams<\/strong>. This method is widely used in chemistry to identify unknown gases when given their mass and volume under standard conditions. It’s especially useful in experimental settings involving gas collection and analysis.<\/p>\n","protected":false},"excerpt":{"rendered":" At STP, 40.0 g of a gas sample takes up 10.3 L. What is the molar mass of this gas? The correct answer and explanation is: Correct Answer: 87.1 g\/mol Explanation: To determine the molar mass of a gas, we need to use the following formula: Molar Mass (g\/mol)=Mass of the gas (g)Moles of the gas (mol)\\text{Molar Mass (g\/mol)} = \\frac{\\text{Mass of the gas (g)}}{\\text{Moles […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-16275","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16275","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=16275"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16275\/revisions"}],"predecessor-version":[{"id":16276,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16275\/revisions\/16276"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16275"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16275"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16275"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nExplanation:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Use the molar volume of a gas at STP<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Calculate the molar mass<\/strong><\/h3>\n\n\n\n
\n\n\n\nConclusion:<\/strong><\/h3>\n\n\n\n