Answer:
The molarity of a 50.0 ml aqueous solution containing 10.0 grams of table sal, Nacl, is 3.42 [tex]\frac{moles}{L}[/tex]
Explanation:
Molarity is a unit of concentration based on the volume of a solution and is defined as the number of moles of solute per liter of solution. Then, the molarity of a solution is calculated by dividing the moles of the solute by the liters of the solution.
[tex]Molarity (M)=\frac{number of moles of the solute}{volume of the solution}[/tex]
Molarity is expressed in units ([tex]\frac{moles}{liter}[/tex]).
Then you must know the amount of moles of the NaCl solute. For that it is necessary to know the molar mass. Being:
the molar mass of NaCl is: 23 g/mole + 35.45 g/mole= 58.45 g/mole
Then a rule of three applies as follows: if 58.45 grams are present in 1 mole of NaCl, 10 grams in how many moles will they be?
[tex]moles=\frac{10 grams*1 mole}{58.45 grams}[/tex]
moles= 0.171
So you know:
Replacing in the definition of molarity:
[tex]Molarity (M)=\frac{0.171 moles}{0.05 L}[/tex]
Solving:
Molarity= 3.42 [tex]\frac{moles}{L}[/tex]
The molarity of a 50.0 ml aqueous solution containing 10.0 grams of table sal, Nacl, is 3.42 [tex]\frac{moles}{L}[/tex]
