The molality of the solution = 17.93 m
<h3>Further explanation</h3>
Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
6 L water = 6 kg water(ρ= 1 kg/L)

Answer:
P = 5.14ATM
Explanation:
Number of moles = 0.108moles
Temperature (T) = 20°C = 20 + 273.15 = 293.15K
Volume V = 0.505L.
Pressure (P) = ?
R = 0.082J/mol.K
From ideal gas equation,
PV = nRT
P = nRT / V
P = (0.108 * 0.082 * 293.15) / 0.505
P = 2.596 / 0.505
P = 5.14ATM
The pressure of the gas is 5.14ATM
2Na + Cl2 ----> 2NaCl
.: product have to be " 2NaCl "