# Chemistry

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(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 -

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 -

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

• Chemistry -

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 -

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.

• Chemistry -

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