What are the number of moles and the molar mass of a gas if 10.22 g of it occupi

Question

What are the number of moles and the molar mass of a gas if 10.22 g of it occupies 5.06 L at 19oC under 0.750 atm? The number of moles is________mol. The molar mass of the gas is _____g/mol. EXPLAIN


The density of carbon dioxide is greater than the density of air and as such it can be used in fire extinguishers to smother the flames by preventing oxygen from the air to feed the fire. What is the density of CO2 at 31oC and 0.98 atm? The density of carbon dioxide under these conditions is_____g/L.  EXPLAIN

What are the number of moles and the molar mass of a gas if 10.22 g of it occupies 5.06 L at 19oC under 0.750 atm? The number of moles is________mol. The molar mass of the gas is _____g/mol. EXPLAIN


The density of carbon dioxide is greater than the density of air and as such it can be used in fire extinguishers to smother the flames by preventing oxygen from the air to feed the fire. What is the density of CO2 at 31oC and 0.98 atm? The density of carbon dioxide under these conditions is_____g/L.  EXPLAIN

The density of carbon dioxide is greater than the density of air and as such it can be used in fire extinguishers to smother the flames by preventing oxygen from the air to feed the fire. What is the density of CO2 at 31oC and 0.98 atm? The density of carbon dioxide under these conditions is_____g/L.  EXPLAIN
The density of carbon dioxide is greater than the density of air and as such it can be used in fire extinguishers to smother the flames by preventing oxygen from the air to feed the fire. What is the density of CO2 at 31oC and 0.98 atm? The density of carbon dioxide under these conditions is_____g/L.  EXPLAIN
The density of carbon dioxide is greater than the density of air and as such it can be used in fire extinguishers to smother the flames by preventing oxygen from the air to feed the fire. What is the density of CO2 at 31oC and 0.98 atm? The density of carbon dioxide under these conditions is_____g/L.  EXPLAIN

Explanation / Answer

Using PV =nRT

0.75 * 5.06 = 10.22/M * 0.0821 * (273 + 19)

M = (10.22 * 0.0821 * 292)/(0.75 * 5.06) = 64.56 g/mol

Hence molar mass of the gas is 64.56 g/mol

No. of moles = 10.22/64.56 = 0.158 mol

As we know PV = nRT

hence PV = m/M RT

P*M = dRT , where M is the molar mass and d is density

0.98 * 44 = d * 0.0821 * 304

d = (0.98 * 44)/(0.0821 * 304) = 1.72 g/L

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