Question

1000g aq. Sol. Of caco3 contains 10g of caco3 conc. Of sol. Is answering in ppm

Answer 1

Answer:

Molar mass of $CaCO_{3}$=100g

Mass of $CaCO_{3}$=10g

Moles of $CaCO_{3}$=mass of solute/molar mass of solute

=10/100

=0.1mol

Mass of solvent=mass of solution-mass of solute

=1000-10

=990g or 0.99kg

Molality = Number of moles of solute / Mass of solvent in Kg

Molality = 0.1/0.99 =  0.1010 M

so convert this into ppm we get 400ppm