Question

Show the calculation of the molarity of a solution made by dissolving 12.6 grams of NaNO3 to make 200 ml of solution.

Answer 1

The molarity of the solution is 0.740 M

Step-by-step explanation:

Step 1: data given

Mass of NaNO3 = 12.6 grams

Molar mass of NaNO3 = 84.99 g/mol

Volume = 200 mL = 0.200 L

Step 2: Calculate moles of NaNO3

Moles NaNO3 = mass NaNO3 / molar mass NaNO3

Moles NaNO3 = 12.6 grams / 84.99 g/mol

Moles NaNO3 = 0.148 moles

Step 3: Calculate molarity

Molarity = moles / volume

Molarity = 0.148 moles / 0.200 L

Molarity = 0.740 M

The molarity of the solution is 0.740 M

Answer 2

Answer: The molarity of a solution made by dissolving 12.6 grams of NaNO3 to make 200 ml of solution is 0.74 M

Step-by-step explanation:

Molarity is defined as the number of moles of solute dissolved perliter of the solution.

`Molarity=(n* 1000)/(V_s)`

where,

n= moles of solute

`V_s` = volume of solution in ml = 1000 ml

`moles of solute =\frac{\text {given mass}}{\text {molar mass}}=(12.6g)/(85g/mol)=0.148moles`

Now put all the given values in the formula of molarity, we get

`Molarity=(0.148moles* 1000)/(200ml)=0.74mole/L`

Thus the molarity of a solution made by dissolving 12.6 grams of NaNO3 to make 200 ml of solution is 0.74 M