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Question

Chemistry Question on Some basic concepts of chemistry

Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is:

A

1.14 mol kg−11.14 \,mol \,kg^{-1}

B

3.28 mol kg−13.28 \,mol \,kg^{-1}

C

2.28 mol kg−12.28 \,mol \,kg^{-1}

D

0.44 mol kg−10.44 \,mol \,kg^{-1}

Answer

2.28 mol kg−12.28 \,mol \,kg^{-1}

Explanation

Solution

molality (m) =M1000 d−MM1×100= \frac{M}{1000 \,d -MM_{1}}\times100 M = Molarity M1M_{1} = Molecular mass d = density =2.05(1000×1.02)−(2.05×60)×1000= \frac{2.05}{\left(1000\times1.02\right)-\left(2.05\times60\right)}\times1000 =2.28 mol kg−1= 2.28 \,mol \,kg^{-1}