0.209 g of a gas occupies 186.6 mL at STP. What is its molecular weight?
回答 (1)
2017-10-30 3:40 pm
Pressure, P = 1 atm
Temperature, T = 273 K
Mass, m = 0.209 g
Molar mass, M = ? g/mol
Volume, V = 186.6 mL = 0.1866 L
Gas constant, R = 0.0821 atm L / (mol K)
PV = nRT and n = m/M
Then PV = (m/M)RT
Hence, M = mRT/(PV)
Molar mass, M = 0.209 × 0.0821 × 273 / (1 × 0.1866) g/mol = 25.1 g/mol
Molecular weight = 25.1
Temperature, T = 273 K
Mass, m = 0.209 g
Molar mass, M = ? g/mol
Volume, V = 186.6 mL = 0.1866 L
Gas constant, R = 0.0821 atm L / (mol K)
PV = nRT and n = m/M
Then PV = (m/M)RT
Hence, M = mRT/(PV)
Molar mass, M = 0.209 × 0.0821 × 273 / (1 × 0.1866) g/mol = 25.1 g/mol
Molecular weight = 25.1
收錄日期: 2021-04-18 17:59:50
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