Question

0.3 g of a gas has a volume of 112 ml at 0°C and 2 atm .its molecular weight is ​

Answer 1

Explanation:

The standard way to do this problem is to use the gas law equation:

PV = nRT

P = pressure = 800/760 = 1.053 atm

V = volume = 0.380L

n = number of moles

R = gas law constant = 0.082057

T = temp in K = 27+273 = 300

Substitute:

1.053* 0.380 = n * 0.082057*300

n = ( 1.053*0.380) / ( 0.082057*300)

n = 0.400/ 24.6171

n = 0.0162 mol

0.455g = 0.0162 mol

1mol = 0.455/ 0.0162 = 28.00g/mol

Answer 2

The molecular weight of the given gas is 30g.

Given:

m = 0.3 g , v = 112 ml t= 0° C = 273Kand p = 2 atm.

To Find:

The molecular weight

Solution:

From ideal gas law

PV = nRT

n= $\frac{m}{w}$

⇒$PV = \frac{m}{w} RT$

$2 * 0.112 = \frac{0.3}{m} *0.082 *273\\\\m = \frac{6.71}{0.224} \\\\m = 29.9$

m ≈ 30 g.

Therefore, the molecular weight of the gas is 30g.

#SPJ3

Learn More

1) An ideal gas has a volume of 15ml, a temperature of 20C and a pressure of 100kPa. The volume of the gas is reduced to 5ml and the temperature is raised to 40C. What is the new pressure of the gas?

Link:https://brainly.in/question/10432043

2)The volume expansion coefficient of an ideal gas Gama at a constant pressure is observed as gamma directly proportional T^n the value of n is

Link:https://brainly.in/question/14502495