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Chemistry Question on Some basic concepts of chemistry

A solution of methanol in water is 20%20\% by volume. If the solution and pure methanol have densities of 0.964  kg  L−10.964 \; kg \; L^{-1} and 0.793  kg  L−10.793 \; kg \; L^{-1}, respectively, find the percent of methanol by weight?

A

15.8

B

16.45

C

20

D

14.8

Answer

16.45

Explanation

Solution

Given, methanol =20 mL=20\,mL

Water =8 mL=8\,mL

density of methanol =0.793 kgL−1=0.793\,kg L ^{-1} or 0.793 gmL−10.793\,gmL ^{-1}

density of solution =0.964 kgL−1=0.964\,kg L ^{-1} or 0.964 gmL−10.964\,gmL ^{-1}

As, density = mass  volume =\frac{\text { mass }}{\text { volume }}
∴\therefore mass/weight of methanol

=0.793 g mL−1×20 mL=15.86 g=0.793\,g \,mL ^{-1} \times 20 \,mL =15.86 \,g

Similarly weight of solution

=0.964 g mL−1×80 mL=96.4 g=0.964\,g \,mL ^{-1} \times 80 \,mL =96.4 \,g

We know that,

%\% by weight = weight of solute  weight of solution ×100=\frac{\text { weight of solute }}{\text { weight of solution }} \times 100
=15.86g96.4 g×100=16.45=\frac{15.86 g }{96.4\,g } \times 100=16.45