#### Thank you for registering.

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

Click to Chat

1800-1023-196

+91 7353221155

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

# NaCl molecule is bound due to electric force between the Sodium and chlorine Ion when one electron of sodium is transferred to chlorine. Taking separation between the ions to be 2.75 * 10^-8 cm, find the force of attraction between them.

Shwetabh
11 Points
3 years ago
r=2.75×10^-8×10^-2 So,r=2.75×10^-10F=1/4piepsnot×q1q2/r^2F=9×10^-9×1.6×10^-19×1.6×10-19/(2.75×10^-10)^2F=9×10^9×1.6×1.6×10^-38×10-20/2.75×2.75F=9×1.6×1.6×10^29-38/2.75×2.75F=9×1.6×1.6×10^-9/2.75×2.75F=23.04/7.56×10^-9So,F=3.04×10-9 NewtonWe have take one one charge of na and cl because it is mentioned in the question that one electron of sodium is transferred to chlorine So, Na=10electron 11 proton Cl=18 electron 17 protonSo,Na and Cl have one charge
Anjali Shukla
13 Points
3 years ago
Let the given separation between the ions be 2.75*10-8 cm
Net charge on chloride ion, q1 = 1.6*10-19 C
Net charge on sodium ion, q2 = 1.6*10-19 C
F= kq1q2/r2(sq.)
F = 9*10 9 * (1.6*10-19)2 / (2.75*10-8)2