Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        The ratio of masses of oxyg​en and nitrogen in a particular gaseous mixture 1:4 the ratio of number of their molecule is.`
4 years ago

noogler
489 Points
```							7:32
```
4 years ago
noogler
489 Points
```							cuz 1 mole of an atom contains 6.023x1023 molecules(N)let x be mass of oxygen ….........4x be mass of nitrogenno. of moles of O2=x/32...............no. of moles of N2=4x/28no.of molecules of O2=(x/32)N...............no. of molecules of N2=(4x/28)Nthen find the ratio of no. of molecules of O2 , N2
```
4 years ago
Supriya vispute
11 Points
```							mass of oxygen/mass of nitrogen=1/4. Mole=weight / atomic weightMole of oxygen be      1x/32 and mole of nitrogen be 4x/32=x/7Mole=no.of molecules /NATherefore,mole of oxygen(x/32)=No.of molecules /NA     Therefore no.of molecules =NA/32.....1Mole of nitrogen (x/7)=No.of molecules /NATherefore NA/7=no.of molecules .......2From1 and 2NA/32=NA/7Therefore 7NA=32NA.....(cross multiplying) i.e.7:32
```
2 years ago
12 Points
```							answer: 7:32iet x be the mass of oxygentherefore , oxygen=x/32similarily,mass of nitrogn is 4xtherefore, nitrogen =4x/28now the ratio =x/32:4x/28=7:32
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Physical Chemistry

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 141 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions