Question

A 0.590 gram sample of a metal, M, reacts completely with sulfuric acid according to:

M(s) +H2SO4(aq) --> MSO4(aq) +H2(g)

A volume of 255 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 Torr and the temperature is 25�C. Calculate the molar mass of the metal.

What is the molar mass of the metal= g/ml

Answer 1

The molar mass of the metal is 59g/mol

Step-by-step explanation:

Step 1: Data given

Mass of the metal = 0.590 grams

volume of hydrogen collected = 255 mL = 0.255 L

Atmospheric pressure = 756.00 torr = 0.9947 atm

vapor pressure of water at 25 °C = 23.8 torr = 0.0313 atm

Temperature = 25.00°C

Step 2: The balanced equation

M(s) +H2SO4(aq) → MSO4(aq) +H2(g)

Step 3: Calculate partial pressure of H2

Only water and hydrogen contributes to the total pressure

ptotal = pH2 + pWater

0.9947 atm = pH2 + 0.0313 atm

pH2 = 0.9947 atm-0.0313 atm = 0.9634 atm

Step 4: Calculate moles of H2:

p(H2)*V = n(H2)*R*T

⇒ p(H2) = partial pressure of H2 = 0.9634 atm

⇒ V = the volume of hydrogen = 255 mL = 0.255 L

⇒ n(H2) = the moles of hydrogen = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 25 °C = 298 Kelvin

n(H2) = (p(H2)*V)/(R*T)

n(H2) = (0.9634 * 0.255)/(0.08206*298)

n(H2) = 0.0100 moles

Step 5: Calculate moles of metal M

For 1 mole H2 produced, we need 1 mole M to be consumed

For 0.0100 moles H2 we have 0.0100 moles M

Step 6: Calculate molar mass of the metal

Molar mass of metal = mass of metal / moles of metal

Molar mass of metal = 0.590 grams / 0.0100 moles

Molar mass of metal = 59 g/mol

The molar mass of the metal is 59g/mol

Answer 2

Molar mass of the metal = 59.0 g/mol

Step-by-step explanation:

Vapor pressure of water at
`25^oC` = 23.78 torr

We are given:

Total vapor pressure = 756.0 torr

Vapor pressure of hydrogen gas = Total vapor pressure - Vapor pressure of water = (756.0 - 23.78) torr = 732.22 torr

To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:

`PV=nRT`

where,

P = pressure of the gas = 732.22 torr

V = Volume of the gas = 255 mL = 0.255 L ( 1 mL = 0.001 L )

T = Temperature of the gas =
`25^oC=[25+273]K=298K`

R = Gas constant =
`62.3637\text{ L.torr }mol^(-1)K^(-1)`

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

`732.22 torr* 0.255L=n* 62.3637\text{ L.torr }mol^(-1)K^(-1)* 298K\\\\n=(732.22* 0.255)/(62.3637* 298)=0.01mol`

From the reaction shown below:-

`M+H_2SO_4\rightarrow MSO_4+H_2`

1 mole of hydrogen gas is produced when 1 mole of the metal is reacted.

Also,

0.01 mole of hydrogen gas is produced when 0.01 mole of the metal is reacted.

Moles of the metal = 0.01 mol

Mass taken = 0.590 g

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`0.01\ mole= (0.590\ g)/(Molar\ mass)`

Molar mass of the metal = 59.0 g/mol