Question
Density of a $$2.05M$$ solution of acetic acid in water is $$1.02g/mL.$$ The molality of the solution is
A.
$$2.28\,{\text{mol}}\,{\text{k}}{{\text{g}}^{ - 1}}$$
B.
$$0.44\,{\text{mol}}\,{\text{k}}{{\text{g}}^{ - 1}}$$
C.
$$1.14\,{\text{mol}}\,{\text{k}}{{\text{g}}^{ - 1}}$$
D.
$$3.28\,{\text{mol}}\,{\text{k}}{{\text{g}}^{ - 1}}$$
Answer :
$$2.28\,{\text{mol}}\,{\text{k}}{{\text{g}}^{ - 1}}$$
Solution :
TIPS/Formulae:
$$\eqalign{
& {\text{Apply the formula }}d = M\left( {\frac{1}{m} + \frac{{{M_2}}}{{1000}}} \right) \cr
& \therefore 1.02 = 2.05\left( {\frac{1}{m} + \frac{{60}}{{1000}}} \right) \cr
& {\text{On solving we get, m = }}2.288{\text{ }}mol/kg \cr} $$