How many moles of of HCl are present in 250cm3 of 0.5 M HCl solution .

Explanation

Mass of NaOH = 4g
Volume of solution = `1/(10)L`
Number of moles of NaOH = `4/(40)=0.1`
`"Molarity"=("No. of moles of solute")/("Volume of solution (L)")=0.1xx1/(1/(10))=0.1xx10=1M`
`"Molality (m)"=("No. of mole of solute")/("Mass of solvent (Kg)")`
Density of solution = `1.038"g mL"^(-1)`
Volume of solution `=1/(10)xx100"mL"=100"mL"`
Mass of solution = `d xxv=1038/(1000)xx1000=103.8g`
Mass of solvent = 103.8 - 4 = 99.8 g
`"Molality"=(0.1xx1000)/(99.8)=1.002m`