Question

Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is 0.826 g/mL. Calculate the concentration of the solution in units of (a) molality; (b) mole fraction of LiBr; (c) mass percentage of CH3CN.

Answer 1

Answer:

(a) $m=2.69m$

(b) $x_{LiBr}=0.099$

(c) $\% LiBr=18.9\%$

Explanation:

Hello,

In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:

(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:

$m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g$

Next, we compute the mass of the solution:

$m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g$

Then, the mass of the solvent (acetonitrile) in kg:

$m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg$

Finally, the molality:

$m=\frac{1.80mol}{0.670kg} \\\\m=2.69m$

(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):

$n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol$

Then, the mole fraction of lithium bromide:

$x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099$

(c) Finally, the mass percentage with the previously computed masses:

$\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%$

Regards.