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
Menu
Grade: upto college level
`        Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. calculate the percentage of each isotope in natural boron.`
5 years ago

## Answers : (2)

Deepak Patra
askIITians Faculty
471 Points
```							Sol. 1. Average atomic weight
= (∑▒〖Percentage of an isotope ×atomic weight〗)/100
⇒ 10.81 = (10.01x+11.01 (100-x))/100
⇒ x = 20%
Therefore, natural boron contain 20% (10.01) isotope and 80% other isotope.

```
5 years ago
jatin
11 Points
```							Let the % of boron with mass no 10 be A. And Mass no 11 be = 100 - A10.A+(100 - A) 11/100= 10.810A + 1100 -11A = 1080-1A = -20A=-20/-1A =20%And for boron with mass no.11=20-100                                                        =80%
```
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## 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

## Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details