Posted by **Anon** on Monday, October 31, 2011 at 1:12am.

(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?

- Chemistry -
**DrBob222**, Monday, October 31, 2011 at 12:01pm
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.

- Chemistry -
**Arg**, Friday, January 13, 2012 at 12:59am
Determine the molarity of a 300.0mL solution that contains 12.05g Na(2)SO(4)

- Chemistry -
**Kate**, Sunday, August 19, 2012 at 2:45pm
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

- Chemistry -
**M**, Sunday, March 16, 2014 at 11:00pm
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.

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