Question

What is the molarity of a 50.0 ml aqueous solution containing 10.0 grams of table sal, Nacl?

Answer 1

The molarity of a 50.0 ml aqueous solution containing 10.0 grams of table sal, Nacl, is 3.42
`(moles)/(L)`

Step-by-step 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.

`Molarity (M)=(number of moles of the solute)/(volume of the solution)`

Molarity is expressed in units (
`(moles)/(liter)`).

Then you must know the amount of moles of the NaCl solute. For that it is necessary to know the molar mass. Being:

• Na: 23 g/mole
• Cl: 35.45 g/mole

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?

`moles=(10 grams*1 mole)/(58.45 grams)`

moles= 0.171

So you know:

• number of moles of solute= 0.171 moles
• volume= 50 mL= 0.05 L

Replacing in the definition of molarity:

`Molarity (M)=(0.171 moles)/(0.05 L)`

Solving:

Molarity= 3.42
`(moles)/(L)`

The molarity of a 50.0 ml aqueous solution containing 10.0 grams of table sal, Nacl, is 3.42
`(moles)/(L)`