Question

What is the molarity of a solution in which 0.45 grams of sodium nitrate are dissolved in 265mL of solution?

Answer 1

Answer:

m=0.45g

as, M.M. of NaNO₃=23+14+48=85g/mol

Given volume, V=265mL=0.265L

As, n=m/M.M.

And, M=n/V

∴ M=m÷(M.M.×V)

⇒M=0.45/(85×0.265)=0.45÷22.525

Therefore, The molarity of a solution in which 0.45 grams of sodium nitrate are dissolved in 265mL of solution is

'0.0199'≈"0.02M"

Answer 2

Answer:

The molarity of the solution is 0.02 M.

Explanation:

• To find the molarity, we need to first find out the number of moles of sodium nitrate in the solution.

The formula to get moles:

$Moles = \frac{Mass}{Molar Mass}$

The mass of sodium nitrate- 0.45g

The molar mass of sodium nitrate- 85g/mol

$Moles = \frac{0.45}{85} = 0.0053$

= 0.0053 mol of sodium nitrate

• Now we can calculate the molarity of sodium nitrate solution.

The formula to calculate molarity:

$Molarity = \frac{Moles}{Volume}$

Moles of sodium nitrate- 0.0053 mol

The volume of the solution- 265 ml = 0.265 L

$Molarity = \frac{0.0053}{0.265} = 0.02$

= 0.02 M of sodium nitrate

Conclusion:

The molarity of the sodium nitrate solution is 0.02 M.