Question

what mass of NaNO3 is required to prepare 400 ml of a 0.45M solution
​

Answer 1

Answer:

Percentage by mass is the amount of a component in a mixture per 100 unit if mass of the total mixture. From the problem, we first need to convert the volume of water to mass by using the density of water of 1000 g/L. We calculate as follows:

Mass of water = .100 L ( 1000 g/L ) = 100 g water

Total mass = 100 g + 20 g = 120 g

Percent mass = 20 / 120 x 100 = 16.67% NaNO3 solution

Explanation:

Hope this helps:)  ジョセフ

Answer 2

Answer:

The mass of $NaNO_{3}$ in 400 ml and 0.45 M solution is 15.3 g.

Explanation:

We are provided the molarity and the volume of the solution.

Molarity - 0.45 M

Volume - 400 ml = 0.4 L

The formula for molarity is:

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

We substitute to values to find moles of $NaNO_{3}$.

$0.45 = \frac{Moles}{0.4}$

$0.45 * 0.4 = 0.18 mol$

We can find the mass of  $NaNO_{3}$ by using the mole formula.

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

The molar mass $NaNO_{3}$ is - 85 g/mol

The moles of $NaNO_{3}$ are - 0.18 mol

$0.18 = \frac{Mass}{85}$

$Mass = 15.3 g$ of $NaNO_{3}$

Conclusion:

The mass of $NaNO_{3}$ is 15.3 g.