Got it here ...

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4091932/

" ... for a given ion, the molarity as reported in simulation literature is based on either the total amount of a specific ion present in the simulation (not just excess salt) or the total number of excess ions per the total volume (converted to moles/liter). Since the system is not a bulk macroscopic system, if the net-neutralizing salt is included in the calculation of molarity, this concentration may be much higher than expected under periodic boundary conditions ..."

Hi Himanshu,

You dont need to add extra Na+ ions in the solution. You need to count the number of Na+ ions present in the box, then the volume of the box. If you know these two things, you can easily calculate the molarity of the Na+ ions in box.

Hope it helps.

Molarity = number of moles (No of particles) / volume (lit)

Hi Himanshu,

You can use this perl script provided by Thomas Cheatham III. The volume and  the molarity is taken as inputs. If the output is say 58 then add 29 Na+ and 29 Cl-

The link for the discussion in AmberMD forum is provided below.

------------------------

molarity.perl

#!/usr/sbin/perl

$molar = $ARGV[0];

$box_x = $ARGV[1];

$box_y = $ARGV[2];

$box_z = $ARGV[3];

$box_a = $ARGV[4];

$box_b = $ARGV[5];

$box_g = $ARGV[6];

if ($box_x == 0) {

printf("Usage: molarity box-x box-y box-z alpha beta gamma\n");

die;

}

if ($box_y == 0) {

$box_y = $box_x;

}

if ($box_z == 0) {

$box_z = $box_x;

}

if ($box_a == 0) {

$box_a = 90.0;

}

if ($box_b == 0) {

$box_b = $box_a;

}

if ($box_g == 0) {

$box_g = $box_a;

}

$torad = 2 * 3.141592654 / 360.0;

$rad_a = $box_a * $torad;

$rad_b = $box_b * $torad;

$rad_g = $box_g * $torad;

$angles = 1 - cos($rad_a)*cos($rad_a) - cos($rad_b)*cos($rad_b) - cos($rad_g)*cos($rad_g);

$angles += 2 * cos($rad_a)*cos($rad_b)*cos($rad_g);

$angles = sqrt($angles);

$volume = $box_x * $box_y * $box_z * $angles;

$molecules = 6.022 * $volume * $molar / 10000;

printf(" MOLARITY = %8.3f\n", $molar);

printf(" Box size = %8.3f %8.3f %8.3f %8.3f %8.3f %8.3f\n", $box_x, $box_y, $box_z, $box_a, $box_b, $box_g);

printf(" Volume = %8.3f\n", $volume);

printf("\n %8.3f molecules are necessary to make a molarity of %6.2f M\n\n", $molecules, $molar);

http://archive.ambermd.org/200907/0332.html

Method 2: requires only the number of water molecules in the box.

N_ions = 0.0187 x ionConc(M) x no. of Water molecules

This discussion related to this is available in VMD forum n the link is provided here.

http://www.ks.uiuc.edu/Research/vmd/mailing_list/vmd-l/3943.html