Calculate the relative molecular mass?
2018-03-04 3:53 pm
when a 0.20 g of a liquid ,y, with a simple molecular structure was evaporated it produced 80 cm^3 of vapour. The temperature was 98 degrees celsius and the pressure 1.1*10^5Pa. Calculate the relative molecular mass of y
回答 (2)
2018-03-04 10:41 pm
For the vapour :
Mass, m = 0.20 g
Pressure, P = 1.1 × 10⁵ Pa
Volume, V = 80 cm³ = 80 × 10⁻⁶ m³
Temperature, T = (273 + 98) K = 371 K
Gas constant, R = 8.314 J / (mol K)
PV = nRT and n = m/M
Then, PV = (m/M)RT
Molar mass, M = mRT/PV = 0.20 × 8.314 × 371 / [(1.1 × 10⁵) × (80 × 10⁻⁶)] g/mol = 70.1 g/mol
Relative molecular mass = 70.1
Mass, m = 0.20 g
Pressure, P = 1.1 × 10⁵ Pa
Volume, V = 80 cm³ = 80 × 10⁻⁶ m³
Temperature, T = (273 + 98) K = 371 K
Gas constant, R = 8.314 J / (mol K)
PV = nRT and n = m/M
Then, PV = (m/M)RT
Molar mass, M = mRT/PV = 0.20 × 8.314 × 371 / [(1.1 × 10⁵) × (80 × 10⁻⁶)] g/mol = 70.1 g/mol
Relative molecular mass = 70.1
2018-03-04 5:29 pm
Use gas equation to calculate mol of Y
PV = nRT
P = pressure = (1.1*10^5Pa) = (1.1*10^5Pa) / (1.01325 ^10^-5 Pa/atm ) = 1.0856 atm
V = 80cm³ = 0.080dm³
N = mol
R = 0.082057
T = 98°C = 371.15K
n = 1.0856*0.080 / ( 0.082057*371.15)
n = 0.00285 mol of Y
0.00285 mol Y has mass = 0.200g
1 mol Y has mass = 0.200/0.00285 = 70.1g/mol
PV = nRT
P = pressure = (1.1*10^5Pa) = (1.1*10^5Pa) / (1.01325 ^10^-5 Pa/atm ) = 1.0856 atm
V = 80cm³ = 0.080dm³
N = mol
R = 0.082057
T = 98°C = 371.15K
n = 1.0856*0.080 / ( 0.082057*371.15)
n = 0.00285 mol of Y
0.00285 mol Y has mass = 0.200g
1 mol Y has mass = 0.200/0.00285 = 70.1g/mol
收錄日期: 2021-04-24 01:06:53
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