Answer: hope this helps
To make molar NaCl solutions of other concentrations dilute the mass of salt to 1000ml of solution as follows:
0.1M NaCl solution requires 0.1 x 58.44 g of NaCl = 5.844g.
0.5M NaCl solution requires 0.5 x 58.44 g of NaCl = 29.22g.
2M NaCl solution requires 2.0 x 58.44 g of NaCl = 116.88g.
Explanation:
Answer:
The molarity of the solution is 0.6 [tex]\frac{moles}{L}[/tex]
Explanation:
Molarity (M) is the number of moles of solute that are dissolved in a given volume.
The Molarity is expressed by:
[tex]Molarity (M)=\frac{number of moles of solute}{dissolution volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
Then you must know the moles that represent 2.5 grams of LiNO₃.
For that you must know the molar mass of the compound. You know the atomic mass of the elements that make up the compound, obtained from the periodic table:
So, taking into account the abundance of each element in the compound:
LiNO₃= 7 g/mol + 14 g/mol + 3*(16 g/mol)= 69 g/mol
Then a rule of three applies as follows: if 69 g of LiNO₃ is 1 mol of the compound, 2.5 g of LiNO₃ how many moles are there?
[tex]moles=\frac{2.5 g*1 mole}{69 g}[/tex]
moles= 0.036
Being 60 mL = 0.06 L (1L = 1000 mL or 1 mL = 0.001 L), the molarity is calculated as:
[tex]M=\frac{0.036moles}{0.06L}[/tex]
M=0.6 [tex]\frac{moles}{L}[/tex]
The molarity of the solution is 0.6 [tex]\frac{moles}{L}[/tex]
