 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

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
```
Calculate density of NaCl if edge length of nacl unit cell is 564 pm

```
2 years ago

```							The volume of a unit cell, Vcell, is given by: Vcell = ZM/dN where Z is the formula units per unit cell M is the molar mass per formula unit d is the density in g cm^-3 N is the Avogadro constant NaCl has a fcc lattice and for fcc, Z = 4 For NaCl, M = 58.44 g mol^-1 As the unit cell is a cube, the edge length 564 pm = 3√Vcell 5.64 x 10^-8 cm = 3√(ZM/dN) (5.64 x 10^-8 cm)^3 = (ZM/dN) dN = ZM/(5.64 x 10^-8 cm)^3 d = [ZM/(5.64 x 10^-8 cm)^3]/N d = [(4 x 58.44 g mol^-1)/(5.64 x 10^-8 cm)^3)]/6.022 x 10^23 mol^-1 d = 2.16 g cm^-3
```
2 years 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