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

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

Original post by fireball359

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

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.

😀A thank you is always appreciated. "A word is enough to the wise."😀

Talk to you soon!

(edited 2 months ago)

Original post by Nitrotoluene

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.

😀A thank you is always appreciated. "A word is enough to the wise."😀

Talk to you soon!

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.

😀A thank you is always appreciated. "A word is enough to the wise."😀

Talk to you soon!

Thank you for your help!

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