A solution is prepared by dissolving 1.0 g of NaOH in water to get 250 ml of solution . Calculate its molarity .

Explanation

Mass of `HNO_(3)` in solution = 68g
Moalr mass of `HNO_(3)=63"g mol"^(-1)`
Mass of solution = 100 g
Density of solution = 1.054 g `ml^(-1)`
Volume of solution = `("Mass of solution")/("Density of solution")`
`=((100g))/((1.504"g ml"^(-1)))=66.5mL=0.0665L`
`"Molarity of solution (M)"=("Mass of "HNO_(3)//"Molar mass of "HNO_(3))/("Volume of solution in litres")=((68g//63"g mol"^(-1))/(0.0665L))`
`=16.23"mol L"^(-1)=16.23M`