A 15.7mL sample of HNO3 reacts to neutralize 27.7mL of 0.187M KOH solution. What is the molarity of the HNO3 solution?

HNO3 + KOH ==> KNO3 + HOH

mols KOH = L x M = 0.0277 x 0.187 = ??
mols HNO3 = mols KOH (1 mol reacts with 1 mol).
mols HNO3 = L x M
You have mols HNO3 and L, solve for M.

Check my thinking. Check my arithmetic.

I lost you on this one, I got a little confused on the 3rd step.

mols = L x M
mols KOH = 0.0277L x 0.187M = 0.00518 mols KOH.
The equation is
HNO3 + KOH ==> KNO3 + HOH

The equation shows 1 mol HNO3 reacts with 1 mol KOH; therefore, 0.00518 mols KOH requires 0.00518 mol HNO3.

Then mols HNO3 = L x M
0.00518 mol HNO3 = L x M
0.00518 mol HNO3 = 0.0157 L x M/
Solve for M.
Check to make sure I didn't make a typo.

Thanks for your help but I cant seem to get this one to come out to any of my possible answers. I think I am just getting confused on this one for some reason.

What are the possible answers?

0.165
0.330
0.498
0.298
0.198

0.0277*0.187/0.0157 = 0.32993 which rounds to 3 s.f. as 0.330. I see one like that. Don't you?

I got it finally.

Thank You

How much of each starting material would you use to prepare 2.00 L of each of the following solutions?
0.400 M potassium chromate from solid potassium chromate?

To determine how much solid potassium chromate you would need to prepare a 0.400 M solution of potassium chromate in 2.00 L, you need to use the formula:

moles = volume (L) x molarity

Rearranging the formula, you can solve for the volume:

volume (L) = moles / molarity

First, you need to determine the number of moles of potassium chromate required. Since the molarity and volume are provided, you can use the formula:

moles = volume (L) x molarity

moles = 2.00 L x 0.400 M = 0.800 moles

Now that you have the moles of potassium chromate needed, you can use its molar mass to calculate the mass required. The molar mass of potassium chromate (K2CrO4) is:

2(39.10 g/mol) + 1(52.00 g/mol) + 4(16.00 g/mol) = 194.20 g/mol

mass (g) = moles x molar mass

mass (g) = 0.800 moles x 194.20 g/mol

mass (g) = 155.36 g

Therefore, to prepare a 0.400 M solution of potassium chromate in 2.00 L, you would need to use 155.36 grams of solid potassium chromate.