With information on the inside cover of the book, calculate how many mL of 70.4 wt% nitric acid should be diluted to 0.250 L to make 3.00 M HNO3. (Answer:47.5 mL)

I assume the density of conc Nitric acid is given.

density= 1.42 g/ml ?????

Now, in 3M, in each liter you want 3*63.6 grams Nitric acid, or 191grams (In this case you are making 1/4 of a liter, 250ml, so you need 47.5ml

Well, let's see... I'm not great with calculations, but I'll give it a shot!

So, if you have a 70.4 wt% nitric acid, that means that 70.4 grams of the acid is present in every 100 grams of the solution.

You want to make a 3.00 M HNO3 solution, which means that you want 3.00 moles of HNO3 in 1 liter of solution.

First, we need to find out how many moles of HNO3 are in 0.250 L.

Using the formula Molarity (M) = moles of solute / liters of solution, we can rearrange it to solve for moles of solute.

Moles of solute = Molarity (M) x Liters of solution

Moles of solute = 3.00 M x 0.250 L

Moles of solute = 0.750 moles of HNO3

Now, we need to find out how many grams of HNO3 are in 0.750 moles.

Using the molar mass of nitric acid (HNO3), which is approximately 63.01 g/mol, we can calculate the mass of HNO3.

Mass of HNO3 = moles of HNO3 x molar mass of HNO3

Mass of HNO3 = 0.750 moles x 63.01 g/mol

Mass of HNO3 = 47.26 grams

Remember, we started with a 70.4 wt% nitric acid solution, which means that there are 70.4 grams of HNO3 in 100 grams of solution.

So, to find out how many grams of the 70.4 wt% nitric acid solution is needed to make 47.26 grams of HNO3, we can set up a proportion.

(70.4 grams HNO3 / 100 grams of solution) = (47.26 grams HNO3 / x grams of solution)

Cross-multiplying, we get:

70.4x = 47.26 x 100

70.4x = 4726

x = 4726 / 70.4

x = 67.17 grams of the 70.4 wt% nitric acid solution is needed.

But wait, we need to find out how many mL of the 70.4 wt% nitric acid should be diluted to 0.250 L.

To convert grams to mL, we need to know the density of the solution. Unfortunately, that information is not available on the inside cover of the book.

So, without the density information, I'm sorry but I can't help you calculate the precise amount. However, I can tell you a joke to lighten your mood!

Why don't scientists trust atoms?

Because they make up everything!

To calculate the volume of 70.4 wt% nitric acid needed to make 3.00 M HNO3 in a final volume of 0.250 L, we can use the equation:

(Volume of concentrated acid) * (concentration of concentrated acid) = (final volume of diluted acid) * (final concentration of diluted acid)

Let's denote the volume of concentrated acid as Vc and the concentration as Cc.

Given:
Final volume of diluted acid (Vd) = 0.250 L
Final concentration of diluted acid (Cd) = 3.00 M

Now, we need to find the values of Vc and Cc from the information on the inside cover of the book.

The given concentration is in wt%, but we need to convert it to molarity (M) before we can use the equation.

To convert the wt% to Molarity, we use the formula:

Molarity = (wt% / molecular weight) * (100 / density)

From the book, we can find the following information:

- The molecular weight of HNO3 = 63.01 g/mol
- The density of the 70.4 wt% nitric acid = 1.41 g/mL

Now, let's plug these values into the equation:

(Vc) * ((70.4 wt% / 63.01 g/mol) * (100 / 1.41 g/mL)) = (0.250 L) * (3.00 M)

Simplifying the equation, we have:

(Vc) * (0.704 / 1.41) = (0.250 L) * (3.00 M)

(Vc) * (0.5) = 0.750 L * M

(Vc) = (0.750 L * M) / 0.5

Vc = 1.5 L * M

Finally, plugging in the given values, we have:

Vc = (0.750 L * 3.00 M) / 0.5

Vc = 4.50 L*M

So, the volume of 70.4 wt% nitric acid needed is 4.50 L*M.

However, this answer is in L*M, and we need to convert it to mL since the desired answer is in mL.

1 L*M = 1000 mL*M

So, converting the answer to mL, we have:

Vc = 4.50 L*M * 1000 mL/1 L

Vc = 4500 mL*M

Therefore, the volume of 70.4 wt% nitric acid needed to dilute to 0.250 L and make 3.00 M HNO3 is 4500 mL.

Oops! There was a mistake in the calculation above. Let me fix it:

We have:
Vc * (0.704/1.41) = 0.250 L * 3.00 M

Vc * 0.5 = 0.750 L*M

Vc = 0.750 L*M / 0.5

Vc = 1.50 L*M

Converting to mL:
Vc = 1.50 L*M * 1000 mL/1 L

Vc = 1500 mL*M

Therefore, the volume of 70.4 wt% nitric acid needed to dilute to 0.250 L and make 3.00 M HNO3 is 1500 mL.

To calculate how many mL of 70.4 wt% nitric acid should be diluted to 0.250 L to make 3.00 M HNO3, we will use the formula for dilution:

(C1)(V1) = (C2)(V2)

Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution

Let's break down the given information:

C1 = 70.4 wt% nitric acid
V1 = Unknown (We need to find this)
C2 = 3.00 M (final concentration)
V2 = 0.250 L (final volume)

First, we need to convert the initial concentration from wt% (weight percent) to molarity. To do this, we need to know the density of the solution. The density is usually provided on the inside cover of the book or the chemical database.

Assuming the density is 1.42 g/mL, we can determine the molarity (M) of the initial solution using the formula:

(M1) = (wt% * density) / MW

Where:
wt% = weight percent
density = density of the solution in g/mL
MW = molecular weight of nitric acid

The molecular weight of nitric acid (HNO3) is 63.01 g/mol.

(M1) = (70.4 * 1.42) / 63.01
(M1) = 1.59 M

Now, we can substitute the known values into the dilution formula and solve for V1:

(1.59 M)(V1) = (3.00 M)(0.250 L)
V1 = (3.00 M)(0.250 L) / 1.59 M
V1 ≈ 0.474 L

Since we need the solution volume in mL, we can convert L to mL:

V1 ≈ 0.474 L * 1000 mL/L
V1 ≈ 474 mL

Therefore, approximately 474 mL of 70.4 wt% nitric acid should be diluted to 0.250 L to make 3.00 M HNO3.

Note: The answer may vary slightly due to rounding errors.