HOW MANY ml OF WATER MUST BE ADDED TO 65ml OF A 5.5 m SOLUTION OF NaOH IN ORDER TO
PREPARE A 1.2-M NaOH SOLUTION
A. 230ml
B. 235ml
C. 229 ml
D. 1L
E. NONE OF THE ABOVE

HOW MANY ml OF WATER MUST BE ADDED TO 65ml OF A 5.5 m SOLUTION OF NaOH IN ORDER TO
PREPARE A 1.2-M NaOH SOLUTION
A. 230ml
B. 235ml
C. 229 ml
D. 1L
E. NONE OF THE ABOVE

using d formula
n1m1v1=n2m2v2
(1)(65)(5.5)=(1)(1.2)(V2)
we get
V2= 65 *5.5/1.2
V2=297 ml( approx)
hence volume of water 2 b added 297-65= 232mlans
(NOT SURE BUT mAYB ITS RIGHT)

I was also thrown by the molality. I tried it M1V1=M2V2 and subtracted but I don't think its correct. Molality is mol/kg (m).

Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

molality = moles of solute / kilograms of solvent

Although their spellings are similar, molarity and molality cannot be interchanged. Compare the molar and molal volumes of 1 mol of solute dissolved in CCl4 (d = 1.59 g / mL)

By definition, a 1 M solution would contain 1 mol of solute in exactly 1.00 L of CCl4 , and a 1 m solution would contain 1 mol of solute in 629 mL of CCl4 .

1 kg of CCl4 x (1000 g / 1 kg) x (mL / 1.59 g) = 629 mL CCl4

uhhh.. noo thats not correct. Look at the kaplan book. Molarity is moles of solute/vol of SOLUTION. this is how i learned it since i was in high school. whatever source u got there is wrong or is not the standard definition.

I see what you mean..... I thought maybe the Kaplan was incorrect (it has been before) I went digging around a little more and found exactly what you guys are saying. I looked it up also in my oooooooooold chem book I had to dig for.

The molality of a solution is calculated by taking the moles of solute and dividing by the kilograms of solvent.
The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.

It is 5.5m not Molarity, I haven't been able to solve the answer. By looking at some of the answer response it is close to the actual answer of 235mL. Choice B. So please explain this problem as clearly as possible. Thanks for the responeses.

5.5 m of NaOH solution = 5.5 moles of NaOH/1000g H20= 5.5 moles of NaOH/55.6mol of H2O. From this, we can say that the mole fraction of H2O in the solution is 55.6/(55.6+5.5)= .91.

So I multiplied 65ml by .91 to get the vol. of water in the 65ml solution.(this is the part im not sure about because the density of NaOH is not given. so this is the only way i could think of). 65ml x.91 gives us 59ml of solvent(H2O).

So now we can use the molality. 59ml of solvent x 5.5 mol NaOH/1000ml solvent (b/c density of water is 1g/1ml) to get .325 mol of NaOH in the 65 ml solution.

And finally, .325mol/(x+.065)L = 1.2 M, x = .206L or 206 mL. Like I said before. This is my best guess.

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