A 0.982-g sample of an unknown gas exerts a pressure of 700 mm Hg in a 450-cm3 container at 23oC. Calculate the molar mass of this gas. [1cm3 = 1mL; 1000mL = 1L]

57.4 g/mol

Explanation:

Step 1: Convert 450 cm³ to L

We will use the conversion factors 1 cm³ = 1 mL and 1000 mL = 1 L.

450 cm³ × 1 mL/1 cm³ × 1 L/1 mL = 0.450 L

Step 2: Convert 23 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 23 + 273.15 = 296 K

Step 3: Calculate the number of moles of the gas

We will use the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 700 mmHg × 0.450 L / (62.4 mmHg.L/mol.K) × 296 K = 0.0171 mol

Step 4: Calculate the molar mass of the gas

0.982 g of the gas correspond to 0.0171 moles.

M = 0.982 g/0.0171 mol = 57.4 g/mol

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

50.23 i believe is correct but don't quote me on that.