How many grams of LiCl are required to make 2.0 L of 0.65 M LiCl solution?
1 answer:
Answer:- 55 g of LiCl are required to make 2.0 L of 0.65 M solution.
Solution:- Given-
molarity of the solution = 0.65M
Volume of solution = 2.0 L
molarity of solution =
It could be arranged as...
moles of solute = molarity x volume of solution
moles of solute = 0.65 x 2.0 = 1.3 moles
Molar mass of LiCl is 42.39 g/mol.
So, mass of LiCl required = 1.3 mol x 42.39g/mol = 55 g
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Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
Explanation :
First we have to calculate the moles of n-butane.
Given:
Molar mass of n-butane = 58.12 g/mole
Mass of n-butane = 58.3 g
Now put all the given values in the above expression, we get:
Now we have to calculate the energy required.
where,
Q = energy required
= enthalpy of fusion of solid n-butane = 4.66 kJ/mol
n = moles = 1.00 mol
Now put all the given values in the above expression, we get:
Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ