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:-
Water is highly ordered. In water each oxygen atom is connected to others around it through hydrogen bonding via bridging hydrogen atoms. When a salt like NaCl is dissolved, some of these Hydrogen bonds break.
When a salt like NaCl dissolves in water, the NaCl breaks in to ions Na+ and Cl-.
The water molecules now surround these ions.
The slightly negative oxygen end of water molecule gets near the Na+, while the slightly positive Hydrogen of water molecule gets near the Cl-.
So before salt sample dissolve, the water molecules were highly ordered due to hydrogen bonding. Now after salt dissolve there is a decrease in order and thus an increase in disorder of the water molecules.
Due to increase in disorder, entropy which is a measure of disorder increases. Since entropy increases, delta S for the process is positive.
Answer : The volume of stock solution needed are, 12.5 mL
Explanation :
Formula used :

where,
are the initial molarity and volume of copper (II) chloride.
are the final molarity and volume of stock solution of copper (II) chloride.
We are given:

Putting values in above equation, we get:

Hence, the volume of stock solution needed are, 12.5 mL
Answer:
see explanation
Explanation:
The reaction has a negative rate law; i.e., Rate = - ΔConcentration / ΔTime which is graphically a negative slope for the plot of Rate as a function of reactant concentration. => Rate ∝ f(Reactant Concentration). However, by raising the temperature, an increase the probability of reaction occurs and the formation of more product.