Question

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

Answer 1

Answer:

Molar mass of the metal = 59.0 g/mol

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:

$732.22 torr\times 0.255L=n\times 62.3637\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=\frac{732.22\times 0.255}{62.3637\times 298}=0.01mol$

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.01\ mole= \frac{0.590\ g}{Molar\ mass}$

Molar mass of the metal = 59.0 g/mol

Answer 2

Answer:

The molar mass of the metal is 59g/mol

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