Home
Class 12
CHEMISTRY
A gaseous mixture contains 1 g of H(2) ,...

A gaseous mixture contains 1 g of `H_(2)` , 4 g of He , 7 g of `N_(2)` and 8 g of `O_(2)` . The gas having the highest partial pressure is :-

A

`H_(2)`

B

`O_(2)`

C

He

D

`N_(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which gas in the mixture has the highest partial pressure, we will follow these steps: ### Step 1: Calculate the number of moles of each gas. To find the number of moles, we use the formula: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \] - For \( H_2 \): \[ \text{Moles of } H_2 = \frac{1 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ moles} \] - For \( He \): \[ \text{Moles of } He = \frac{4 \text{ g}}{4 \text{ g/mol}} = 1 \text{ mole} \] - For \( N_2 \): \[ \text{Moles of } N_2 = \frac{7 \text{ g}}{28 \text{ g/mol}} = 0.25 \text{ moles} \] - For \( O_2 \): \[ \text{Moles of } O_2 = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ moles} \] ### Step 2: Calculate the total number of moles in the mixture. \[ \text{Total moles} = \text{Moles of } H_2 + \text{Moles of } He + \text{Moles of } N_2 + \text{Moles of } O_2 \] \[ \text{Total moles} = 0.5 + 1 + 0.25 + 0.25 = 2 \text{ moles} \] ### Step 3: Calculate the mole fraction of each gas. The mole fraction (\(X\)) is given by: \[ X = \frac{\text{Moles of gas}}{\text{Total moles}} \] - For \( H_2 \): \[ X_{H_2} = \frac{0.5}{2} = 0.25 \] - For \( He \): \[ X_{He} = \frac{1}{2} = 0.5 \] - For \( N_2 \): \[ X_{N_2} = \frac{0.25}{2} = 0.125 \] - For \( O_2 \): \[ X_{O_2} = \frac{0.25}{2} = 0.125 \] ### Step 4: Determine the partial pressure of each gas. The partial pressure (\(P_i\)) can be calculated using: \[ P_i = X_i \times P_{\text{total}} \] Assuming \(P_{\text{total}} = 1\) atm for simplicity (the actual value will not change the comparison): - For \( H_2 \): \[ P_{H_2} = 0.25 \times P_{\text{total}} = 0.25 \text{ atm} \] - For \( He \): \[ P_{He} = 0.5 \times P_{\text{total}} = 0.5 \text{ atm} \] - For \( N_2 \): \[ P_{N_2} = 0.125 \times P_{\text{total}} = 0.125 \text{ atm} \] - For \( O_2 \): \[ P_{O_2} = 0.125 \times P_{\text{total}} = 0.125 \text{ atm} \] ### Conclusion: The gas with the highest partial pressure is **Helium (He)**, with a partial pressure of \(0.5 \text{ atm}\).
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • BASIC INTRODUCTION AND NOMENCLATURE

    ALLEN|Exercise MCQ|12 Videos
  • Chemical bonding

    ALLEN|Exercise All Questions|39 Videos

Similar Questions

Explore conceptually related problems

A gasesous mixture contains 56 g N_(2) ,44 g CO_(2) and 16 g CH_(4) . The total pressure of the mixture is 720 mm Hg. What is the partial pressure of CH_(4) ?

A gaseous mixture containing 8g of O_(2) and 227 mL of N_(2) at STPis enclosed in flask of 5 L capacity at 0^(@)C . Find the partial pressure of each gas and calculate the total pressure in the vessel.

Knowledge Check

  • A gaseous mixture contains 56 g of N_(2), 44 g of CO_(2) and 16 g of CH_(4) The total pressure of the mixture is 720 mm Hg. The partial pressure of CH_(4) is:

    A
    180 mm
    B
    360 mm
    C
    540 mm
    D
    720 mm
  • A gaseous mixture contains 56 g of N_(2) 44 g CO_(2) and 16 g of CH_(4) . The total pressure of the mixture is 720 mm Hg. The partial pressure of CH_(4) is

    A
    180 mm
    B
    360 mm
    C
    540 mm
    D
    720 mm
  • Gaseous mixture of contains 56 g of N_(2) , 44g of CO_(2) and 16 g of CH_(4) . The total pressure of mixture is 720 mm of Hg . The partial pressure of CH_(4) is :-

    A
    75 mm
    B
    160 mm
    C
    180 mm
    D
    215 mm
  • Similar Questions

    Explore conceptually related problems

    A gaseous mixture contains 56 g of N_2 , 44 g of CO_2 and 16 g of CH_4 . The total pressure of the mixture is 720 mm of Hg. The partial pressure of CH_4 is

    A gaseous mixture contains 56 g of N_2 , 44 g CO_2 and 16 g of CH_4 The total pressure of the mixture is 720 mm Hg. The partial pressure of CH_4 is

    A gaseous mixture contains 4.0g of H_(2) and 56.0g of N_(2) . The mole fraction of H_(2) in the mixture is

    The mole of fraction of nitrogen, in a mixture of 7 g of N_(2) and 16 g of O_(2) is

    A vessel of volume 4 litres contains a mixture of 8g of O_(2), 14g of N_(2) and 22 g of CO_(2) at 27^(0)C . The pressure exerted by the mixture is