Question

Calculate the molarity of a solution made by putting 55.8 g of nano3 into a beaker and diluting to 2.50 l.

Answer 1

If the molar weight of NaNO3 is 85g/mol, then the amount of molecule in this question would be: 55.8g/ (85g/mol)= 0.656 moles. The NaNO3 is solved by 2.5 l of water. Molarity is a unit of concentration of a solution that using unit moles of molecule per liter. The concentration/molarity of the solution would be:

concentration= number of moles / volume
concentration= 0.656 moles/ 2.5l= 0.26 molar

Answer 2

Answer : The molarity of a solution is 0.263 mole/L.

Solution : Given,

Mass of $NaNO_3$ = 55.8 g

Volume of solution = 2.50 L

Molar mass of $NaNO_3$ = 84.99 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

$\text{ number of moles solute}=\frac{\text{ Given mass of solute}}{\text{ Molar mass of solute}}$

$Molarity=\frac{w}{MV}$

where,

w = mass of solute

M = molar mass of solute

V = volume of solution

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

$Molarity=\frac{55.8g}{(84.99g/mole)\times (2.50L)}=0.263mole/L$

Therefore, the molarity of a solution is 0.263 mole/L.