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C [mol/L] = w% * rho [g/mL] *1000 [mL/L] / (100% * M [g/mol] ) =

= w% * rho [g/mL] *10 [mL/L%] / ( M [g/mol] ) =

= 15.00 % * 1.04 [g/mL] *10 [mL/L%] / ( 158.034 [g/mol] ) =

= 0.987 [mol/L]

the molarity of the solution of potassium [in mol per liter] equals the content of KMnO4 by mass in percents

multiply by density in g/mL

multiply by 10 [mL/(L*%)]

divide by molar mass of KMnO4 (M = 158.034 g/mol)

This gives 0.987 mol/L. Thus the molarity equals approximately 1 M.

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Q: Calculate the molarity of a solution of potassium permanganate and density is 1.04 gmL which contans 15.00 percent by mass?
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