A laboratory procedure calls for making 400.0 mL of a 1.1 M NnO3 solution. What mass of NaNO3 (in g) is needed?

All you need to do is to substitute your problem numbers into the equations I've given you.

M = moles/L
1.1 = moles/0.400 L
Solve for moles, the only unknown.
moles = 1.1 x 0.400 = 0.440 mols.

Second equation:
moles = grams/molar mass
moles from above = 0.440
molar mass NaNO3 is about 85 (23+14+3*16) = about 85 but you need to confirm that.
So 0.440 = grams/85
Solve for grams, the only unknown.
g = 0.440 x 85 =37.4 grams NaNO3 in 400 mL (0.400 L).

please

M = moles/L

Solve for moles.

moles = grams/molar mass
solve for grams.

Well, it looks like we have a "salting" mission here! Don't worry, Na-thing is impossible for Clown Bot!

To solve this problem, we need to use the formula:

Molarity (M) = moles of solute / liters of solution

First, let's convert 400.0 mL to liters. That's equivalent to 0.4 L.

We can rearrange the formula to solve for moles of solute:

moles of solute = Molarity (M) x liters of solution

So, to find the moles of NaNO3, we multiply the molarity (1.1 M) by the liters of solution (0.4 L):

moles of NaNO3 = 1.1 M x 0.4 L

Now that we have the moles of NaNO3, we can find the mass of NaNO3 using its molar mass. The molar mass of NaNO3 is approximately 85 g/mol.

mass of NaNO3 = moles of NaNO3 x molar mass of NaNO3

mass of NaNO3 = (1.1 M x 0.4 L) x 85 g/mol

Now it's calculation time! Multiply those numbers and let me know the result.

To find the mass of NaNO3 needed to make a 1.1 M solution, you need to calculate the number of moles required and then use the molar mass of NaNO3 to convert it to grams.

Here's how you can do it step by step:

1. Start by using the formula for molarity (M):

Molarity (M) = Moles of solute / Volume of solution (in liters)

Rearrange the formula to find the moles of solute:

Moles of solute = Molarity (M) × Volume of solution (in liters)

Convert the given volume of the solution from milliliters (mL) to liters:

Volume of solution = 400.0 mL ÷ 1000 = 0.4 L (since 1 L = 1000 mL)

2. Substitute the values into the formula:

Moles of solute = 1.1 M × 0.4 L = 0.44 moles

3. Now, find the molar mass of NaNO3:

NaNO3: Na = 1 × 23 g/mol
N = 1 × 14 g/mol
O = 3 × 16 g/mol

Molar mass of NaNO3 = 23 + 14 + (3 × 16) = 85 g/mol

4. Finally, calculate the mass of NaNO3 needed:

Mass of NaNO3 = Moles of NaNO3 × Molar mass of NaNO3

Mass of NaNO3 = 0.44 moles × 85 g/mol ≈ 37.4 g

Therefore, approximately 37.4 grams of NaNO3 is needed to make 400.0 mL of a 1.1 M NaNO3 solution.

maybe poopoo

I dunno......I hate chemistry!