Since you have not mentioned whether the 1:3 cement sand mortar is as per weight batching or volume batching, here I'll describe the process for both the types.

Weight batching:

- Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 and suppose w/c ratio is 0.4
- This means of 1 kg mortar, 0.4 part is water,
- 1 part is cement and 3 parts are sand. 1 part = 1 kg /( 0.4 + 1 + 3)
- This gives, 1 part = 0.227

Quantities of different constituents are,

- water = 0.4 * 1 part =0.4 * 0.227 kg = 0.09 kg
- Cement = 1 part = 0.227 kg
- Sand = 3* 1 part = 0.683 kg

If the sand used for this mix contained adsorbed water equal to 5% of weight of sand, the calculations would change as follows:

- Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 and suppose w/c ratio is 0.4
- This means of 1 kg mortar, 0.4 part is water, 1 part is cement and 3 parts are sand.
- The dry sand weighs only 1/1.05 times ie: .952 times of moist sand. Hence, to maintain 3 parts dry sand, moist sand will have to be added in 3*1.05 part = 3.15 part.
- Of the total 0.4 part water required in the mortar, 5% * 3 part = .15 part will be contributed from adsorbed water in sand and only 0.25 part water will be required to be added externally.
- 1 part = 1 kg /( 0.25 + 1 + 3 + .15)
- This gives, 1 part = 0.227

Quantities of different constituents are,

- water required = 0.25 * 1 part =0.25 * 0.227 kg = 0.056 kg
- Cement = 1 part = 0.227 kg
- Sand = 3.15* 1 part = 0.717 kg

Volume batching:

- Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 by volume and suppose w/c ratio is 0.4
- Assume bulk densities of water as 1 g/c, cement as 1.4 g/cc and fine aggregate as 1.6 g/cc.
- So, mass ratio of cement: sand is 1 part * 1.4 : 3 part * 1.6 .
- This means of 1 kg mortar, 0.4 part is water,
- 1 part is cement and 3.43 parts are sand. 1 part = 1 kg /( 0.4 + 1 + 3.43)
- This gives, 1 part = 0.207

Quantities of different constituents are,

- water = 0.4 * 1 part =0.4 * 0.207 kg = 0.083 kg = 0.83 L of water.
- Cement = 1 part = 0.207 kg = 0.207/ 1.4 = 0.1478 L of cement
- Sand = 3.43* 1 part = 0.71 kg = 0.71/1.6 = 0.444 L of dry sand.

If the sand used for this mix contained adsorbed water equal to 5% of weight of sand, which also caused bulking of sand volume by 20% the calculations would change as follows

- Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 by volume and suppose w/c ratio is 0.4
- Assume bulk densities of water as 1 g/c, cement as 1.4 g/cc and fine aggregate as 1.6 g/cc.
- So, mass ratio of cement: sand is 1 part * 1.4 : 3 part * 1.6 .
- This means of 1 kg mortar, 0.4 part is water, 1 part is cement and 3.43 parts are sand.
- Of the total 0.4 part water required in the mortar, 5% * 3.43 part = .172 part will be contributed from adsorbed water in sand and only 0.228 part water will be required to be added externally. 1 part = 1 kg /( 0.4 + 1 + 3.43)
- This gives, 1 part = 0.207

Quantities of different constituents are,

water = 0.4 * 1 part =0.4 * 0.207 kg = 0.083 kg = 0.83 L of water.

Cement = 1 part = 0.207 kg = 0.207/ 1.4 = 0.1478 L of cement

Changed bulk density of sand is 1.6 * 1.05/1.2 = 1.4 ( because of both adsorbed

Sand = 3.43* 1 part = 0.71 kg = 0.71/1.4 = 0.50 L of moist sand.