Question

Acetonitrile, \[CH_{3}CN\], Is a polar organic solvent that dissolver a wide range of solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is \[0.826 \frac{g}{cm^{3}}\].

Calculate the concentration in:
(a) molality
(b) Mole fraction of LiBr
(c) mass percentage of \[CH_{3}CN\]

Answer 1

You're given the density of acetonitrile?

Answer 2

let me check..

Answer 3

not its density is 0,786 according to wikipedia. my first guess is that 0.826 is the density of the solution..

Answer 4

My first thought was to convert this to grams/Liter of solution,

\[\frac{ cm^{3} }{ 1000 } = L \]

\[0.826*(\frac{ g }{ \cancel\cm^{3} })*(\frac{ \cancel\cm^{3} }{ 1000~L }) \ = \frac{ 0.826g }{ 1000L} =0.000826 (g/L)\]

Answer 5

(8.26 *10 ^-4) -( 1.80mol of LiBr)(87 g/mol of Li Br ) = grams of Acetonitrile

Answer 6

if that's is right i think i can go on by my own...:)

Answer 7

hey @rock_mit182 that's the density given for the LiBr solution right?

Answer 8

I think so...

Answer 9

@Cuanchi @aaronq

Answer 10

@Kainui hey man do you have a minute?

Answer 11

@rock_mit182 yes that is fine,
then
1000g solvent x (156.3 g LiBr /669.6 g solvent) = 233.4g LiBr

233.4 g LiBr x (1 mol LiBr / 86.845 g) = 2.68 molal

molar fraction of LiBr = moles of LiBr / total moles

156.3/86.845= 1.8 moles of LiBr
669.6/ 41.05 = 16.13 moles of acetonitrile

X LiBr = 1.8/ (1.8+16.13)= ????

mass percent = mass of LiBr x 100 / total mass of solution

mass percent = 156.3g x 100 / ( 156.3g + 669.6 g) = ??????

Answer 12

Thanks to all of you :)