Last Question---

[ManzRDH]

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How many ml of H20 must be added to 65 ml of a 5.5 m solution of NaOH in order to perpare a 1.2 M NaOH solution?

a- 230 ml
b- 235 ml
c- 229 ml
d- 1 L
e- none

 

[rsweeney]

How many ml of H20 must be added to 65 ml of a 5.5 m solution of NaOH in order to perpare a 1.2 M NaOH solution?

a- 230 ml
b- 235 ml
c- 229 ml
d- 1 L
e- none

I would say none of the above:

5.5m/(0.065L + X1L) = 1.2M

Thus, 5.5m = 1.2M(0.065L + X1) --> 5.5m = 0.078m + 1.2MX1
Thus, 5.422m = 1.2MX1
Thus, X1 = ~ 4.51L

If I am totally wrong I apologize. I figured I would give it a shot. I want to say M1V1 = M2V2 needs to be applied here.

 

[dexadental]

m1v1=m2v2 is applied here.

(5.5 moles/L)*.065L=(1.2moles/L)*v2
What is v2? Do the math...
= .298 L which equals about 299 mL, I'm thinking its none of the above.
Just a quick look at this problem, don't have time to do the details...I think its right though.

 

[lakersfan]

m1v1=m2v2 is applied here.

(5.5 moles/L)*.065L=(1.2moles/L)*v2
What is v2? Do the math...
= .298 L which equals about 299 mL, I'm thinking its none of the above.
Just a quick look at this problem, don't have time to do the details...I think its right though.

This is right but I think there is one more step to the problem. The question asks how much water should be added. Since V2=298 ml and we already had 65 ml (v1), the volume of H20 that should be added is 298-65=233ml.

I think that should be answer...if I did it wrong, please make sure to let the OP know.

 

[luder98]

First thing you need to pay attention to is that the problem gives you molality (m) not molarity (M). So you need to convert it to molarity first. Fortunately, the solvent is water in this case, so the two are the same. You can just use the formula M1V1 = M2V2

65*5.5 = (65+V)*1.2 or V = 233mL

 

[dexadental]

My apologies, yes I agree with the former, I was wrong...damn molality.

 

[ManzRDH]

Thanks!! 🙂