Question

Solve this :

$$\begin{gathered} Giventhat10gofadibasicacid\left(molarmass=100\right)arepresentin600mLofthesolution.Thedensityofthesolutionis1.02gm{L}^{-1}.Molarityofsolutionis \\ \left(1\right)0.17 \\ \left(2\right)0.34 \\ \left(3\right)0.99 \\ \left(4\right)0.016 \end{gathered}$$

Answer 1

Dear Student,

The computation is expressed below,
$$\begin{gathered} dibasicacid=\frac{{\displaystyle \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}}{0.6}=0.17M \\ density=1.02\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$ml$}\right. \\ weightofdibasicacid=10 \\ weightofsolution=600\times 1.02=612 \\ weightofwater=612-10=602g \\ Molarity=\frac{0.17\times 600}{602}=0.169\approx 0.17 \\ AnswerisOption\left(1\right)0.17 \end{gathered}$$

Regards.