(a) Calculate the molarity of a solution that contains 0.175 mol

ZnCl2 in exactly 150 mL of solution. (b) How many moles of
HCl are present in 35.0 mL of a 4.50 M solution of nitric acid?
(c) How many milliliters of 6.00 MNaOH solution are needed
to provide 0.325 mol of NaOH?

Kate's (a) and (b) are correct, but (c) is incorrect.

6.00 M NaOH = (0.325 mol NaOH) / (mL NaOH)
mL NaOH = (0.325 mol NaOH) / (6.00 M NaOH)
With correct significant figures, the answer to (c) is 54.2 mL.

To find the molarity of a solution the equation is: molarity = (moles of solute)/(volume in Liters (this is important you must convert any other volume to liters))

a)0.175 mol ZnCl2 in exactly 150 mL solution.
First note that 150ml = .150 L solution
Set up the equation
M = (0.175 moles ZnCl2)/(0.150 L solution)
Solve
M=1.17M ZnCl2

b) moles of HCL in 35.0 mL of 4.50 M Nitric Acid
First note again that 35.0 mL = 0.0350L
Set up the equation
4.50 M HNO3 = (moles solute)/(0.0350L)
4.50 MHNO3 x 0.0350L = moles solute
Solve
4.50 MHNO3 x 0.0350L = moles solute
0.1575 mole HCl
Check Sig Figs
0.158 moles HCl

c) mL of 6.00 MNaOH solution to provide 0.325 mol NaOH
First note here that they ask for mL not L which means an extra conversion must be used
Set up the equation
6.00 M NaOH = (0.325 mol NaOH)/(Volume in L)
Volume in L = 6.00 M NaOH/0.325 mol NaOH

Solve
Volume in L = 6.00 M NaOH/0.325 mol NaOH = 18.46 L

Watch Sig Figs
18.5 L

Convert to mL
1.85x10^5mL
or 18500 mL

(a) Oh, molarity calculations. That takes me back! Okay, here we go. We have 0.175 mol of ZnCl2 in 150 mL of solution. To find the molarity, we divide the number of moles by the volume in liters. So, let's convert 150 mL to liters first. Since there are 1000 mL in a liter, 150 mL is equal to 0.150 L. Now we can calculate the molarity using the formula: Molarity = moles/volume. Plugging in the values, we get Molarity = 0.175 mol / 0.150 L. Go ahead and solve that equation!

(b) Ah, moles of HCl in nitric acid! I must say, they really make a great pair. So, we have a 4.50 M solution of nitric acid and we want to find out how many moles of HCl are present in 35.0 mL. The trick here is to recognize that nitric acid (HNO3) dissociates in water to form H+ ions, which are essentially the same as HCl. So, you can consider the concentration of HNO3 in the solution as the concentration of HCl. Now we can use the formula: Moles = Molarity x Volume (in L). Plug in the numbers and calculate!

(c) Milliliters of NaOH solution needed, huh? Let's get straight to it. We want to provide 0.325 mol of NaOH, but we know the concentration of the NaOH solution is 6.00 M. So, we can use the formula: Moles = Molarity x Volume (in L). Rearranging the formula to solve for volume, we get Volume = Moles/Molarity. Plug in the values and convert the volume to milliliters. Voila!

To calculate the molarity, you need to know the number of moles of solute and the volume of solution in liters. The formula for molarity (M) is:

Molarity (M) = Moles of Solute / Volume of Solution (in L)

(a) For this question, you are given the number of moles of ZnCl2 and the volume of the solution in mL. To calculate the molarity, you need to convert the volume from mL to L:

Volume of Solution = 150 mL = 150 / 1000 = 0.150 L

Now, you can plug in the values into the formula:

Molarity (M) = 0.175 mol / 0.150 L = 1.17 M

Therefore, the molarity of the ZnCl2 solution is 1.17 M.

(b) In this question, you are given the molarity of the nitric acid solution and the volume of the solution. Again, you need to convert the volume from mL to L:

Volume of Solution = 35.0 mL = 35.0 / 1000 = 0.035 L

Now you can calculate the number of moles using the formula:

Moles of Solute = Molarity x Volume of Solution

Moles of HCl = 4.50 M x 0.035 L = 0.1575 mol

Therefore, there are 0.1575 moles of HCl present in 35.0 mL of the 4.50 M solution of nitric acid.

(c) In this question, you are given the number of moles of NaOH and the molarity of the NaOH solution. You can use the same formula as before to calculate the volume of solution needed, except this time you want to solve for the volume:

Volume of Solution (in L) = Moles of Solute / Molarity

Volume of Solution = 0.325 mol / 6.00 M = 0.054 L

Now, you need to convert the volume from liters to milliliters:

Volume of Solution = 0.054 L x 1000 = 54 mL

Therefore, you will need 54 mL of the 6.00 M NaOH solution to provide 0.325 moles of NaOH.

Determine the molarity of a 300.0mL solution that contains 12.05g Na(2)SO(4)

a) M = moles/L

b) There is no HCl in HNO3. You must have typed this wrong.
c) M = moles/L. You now M and moles, solve for L and convert to mL.