A) 0.1 M
B) 0.05 M
C) 0.2 M
D) 0.01 M
E) None of these
Correct Answer: B
Solution :
Amount of \[{{H}_{2}}S{{O}_{4}}=9.8\,g\] Volume of solution = 2 L Moles of \[{{H}_{2}}S{{O}_{4}}=\frac{\text{mass}\,\text{of}\,{{H}_{2}}S{{O}_{4}}}{\text{molar}\,\text{mass}}\] \[=\frac{9.8\,g}{98\,g\,mo{{l}^{-1}}}\] (\[\because \] Molar mass of\[{{H}_{2}}S{{O}_{4}}\]= 98) Molarity\[=\frac{\text{moles}\,\text{of}\,{{H}_{2}}S{{O}_{4}}}{\text{volume}\,\text{of}\,\text{solution}}\] \[=\frac{0.1\,\,mol}{2\,L}\] \[=0.05\,mol\,{{L}^{-1}}\]or \[0.05\,M\]
You need to login to perform this action.
You will be redirected in
3 sec
