Home
Class 12
CHEMISTRY
The mass of 2.24 xx 10^(-3) m^(3) of a g...

The mass of `2.24 xx 10^(-3) m^(3)` of a gas is 4.4 g at 273.15 K and 101.325 kPa pressure. The gas may be

A

NO

B

`NO_(2)`

C

`C_(3) H_(8)`

D

`NH_(3)`

Text Solution

Verified by Experts

The correct Answer is:
C

m = 4.4 g , V = `2.24 xx 10^(-3) = 2.24 L`
T = 273.15 K
p = 101. 325kPa
= 101325Pa = 1 atm
pV = nRT or pV `= ( m )/( M ) RT `
or `M = ( m RT )/( pV )`
` = ( 4.4 xx 0.0821 xx 273.15)/( 1 xx 2.24 ) = 44.05 `
`:.` The gas is `C_(3) H_(8)`
Promotional Banner

Topper's Solved these Questions

  • STATES OF MATTER

    MODERN PUBLICATION|Exercise MULTIPLE CHOICE QUESTIONS (LEVEL-II)|95 Videos
  • STATES OF MATTER

    MODERN PUBLICATION|Exercise MULTIPLE CHOICE QUESTIONS (LEVEL-III)|7 Videos
  • SOME BASIC CONCEPTS OF CHEMISTRY

    MODERN PUBLICATION|Exercise RECENT EXAMINATION QUESTION|15 Videos
  • STRUCTURE OF ATOM

    MODERN PUBLICATION|Exercise RECENT EXAMINATION QUESTIONS|11 Videos

Similar Questions

Explore conceptually related problems

The mass of 112 cm^(3) " of " CH_(4) gas at STP is

The density of a gas is 1.964 g dm^(-3) at 273 K and 76 cm Hg. The gas is

In the determination of molecular mass by Victor - Meyer's Method 0.790 g of a volatile liquid displaced 1.696 xx 10^(-4) m^(3) of moist air at 303 K and at 1 xx 10^(5) Nm^(-2) pressure. Aqueous tension at 303 K is 4.242 xx 10^(3) Nm^(-2) . Calculate the molecular mass and vapour density of the compound .

If 22.4 ml of a triatomic gas has a mass of 0.048 g at 273K and 1 atm pressure, then the mass of one atom is:

10 dm^(3) of N_(2) gas and 10 dm^(3) of gas X at the same temperature and pressure contain the same number of molecules. The gas X is

Henry 's law constant for the solubility of N_(2) gas in water at 298 K is 1.0 xx 10^(5) atm . The mole fraction of N_(2) in air is 0.6 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm pressure is a. 3.0 xx 10^(-4) b. 4.0 xx 10^(-5) c. 5.0 xx 10^(-4) d. 6.0 xx 10^(-6)