Question

What weight of sodium nitrate (NaNO3) must be used to prepare 506mL of a 10.5 M solution of this salt in water? Answer in units of g.

Answer 1

The question requires us to calculate the mass of NaNO3 necessary to prepare 506 mL of a 10.5 M solution of this salt.

To answer this question, we'll need to go through the following steps:

1) Calculate the molar mass of NaNO3, as it will be necessary during the question;

2) Calculate the number of moles necessary to prepare 506 mL of the 10.5 M solution;

3) Convert the number of moles calculated into the mass of NaNO3, using the molar mass of this salt.

Next, we'll follow the steps:

1) To calculate the molar mass of NaNO3, we need to consider the atomic masses of Na, N and O and the number of atoms of each of these elements. The atomic masses are 22.99 u, 14.01 u and 15.99 u for Na, N and O, respectively. With this information, we can calculate the molar mass:

molar mass of NaNO3 = (1 * 22.99) + (1 * 14.01) + (3 * 15.99) = 84.97 g/mol

2) Next, we use the molar concentration provided (10.5 M or 10.5 mol/L) and the volume required (506 mL = 0.506 L) to obtain the number of moles necessary to prepare this solution:

1 L of solution --------------- 10.5 mol of NaNO3

0.506 L of solution -------- x

Solving for x, we have:

$x=\frac{(0.506\text{ L of solution)}\times(10.5\text{ mol of NaNO}_3)}{(1\text{ L of solution)}}=5.31\text{ mol of NaNO}_3$

Therefore, the number of moles of NaNO3 necessary to prepare the solution is 5.31 moles.

3) At last, we use the molar mass obtained in step 1 (84.97 g/mol) to calculate the mass that corresponds to 5.31 moles of NaNO3:

1 mol of NaNO3 ---------------- 84.97 g

5.31 mol of NaNO3 ----------- y

Solving for y, we have:

$y=\frac{(5.31\text{ mol of NaNO}_3)\times(84.97\text{ g of NaNO}_3)}{(1\text{ mol of NaNO}_3)}=451.2\text{ g of NaNO}_3$

Therefore, the mass of NaNO3 necessary to prepare the solution given by the question is 451.2 g.