chem

Which one of the following contains the greatest number of moles of methanol? (The Avogadro number (L) is 6.02 × 1023, the relative molecular mass (Mr) of methanol is 32.)
A 6.6 × 1022 molecules
B 3.3 g of methanol
C 2.5 × 10−3 m3 of methanol vapour at 300 K and 100 kPa
D 70 cm3 of 1.5 M aqueous methano

Hello fireball!
I'm happy to help you with your problem. However, this approach won't help you learn chemistry effectively.
A. 6.6 × 10^22 molecules
To find the number of moles, you need to divide the number of molecules by Avogadro's number (L):
6.6 × 10^22 molecules / 6.02 × 10^23 molecules/mol = 0.109 mol
B. 3.3 g of methanol
You need to find the number of moles of methanol in 3.3 g:
Molar mass of methanol (Mr) = 32 g/mol ==> Number of moles = mass/molar mass = 3.3 g / 32 g/mol = 0.103 mol
C. 2.5 × 10^−3 m^3 of methanol vapour at 300 K and 100 kPa
To find the number of moles, you need to use the ideal gas equation:
PV = nRT
where P is the pressure (100 kPa), V is the volume (2.5 × 10^−3 m^3), R is the gas constant (0.08206 Lxatm/molxK), and T is the temperature (300 K).
You need to convert the pressure from kPa to atm:
100 kPa × (1 atm / 101.325 kPa) = 0.988 atm
Now, you can put in the values:
(0.988 atm) × (2.5 × 10^−3 m^3) = n × (0.08206 Lxatm/molxK) × (300 K)
Solving for n, you get:
n = 0.019 mol
D. 70 cm^3 of 1.5 M aqueous methanol
You need to convert the volume from cm^3 to L:
70 cm^3 × (1 L / 1000 cm^3) = 0.07 L
The concentration of methanol is 1.5 M, which means there is 1.5 moles of methanol per litre of solution. To find the number of moles in 0.07 L, you have to multiply the concentration by the volume:
1.5 M × 0.07 L = 0.105 mol
You have to compare the number of moles for each option:
A: 0.109 mol
B: 0.103 mol
C: 0.019 mol
D: 0.105 mol
The greatest number of moles of methanol is in option A, which contains 0.109 mol.
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