Question

40% w/v of NaCl solution (specific gravity = 1.12) is equivalent to:

Answer 1

40% (w/v) of NaCl solution, means 40g of NaCl is present in 100ml of solution.

number of mole of NaCl = 40/58.5

given, specific gravity = 1.12

we know, specific gravity is nothing, just relative density of a substance.

so, relative density of solution = 1.12

or, density of solution/density of water at 4°C = 1.12

or, density of solution / 1g/ml = 1.12

or, density of solution = 1.12 g/ml

mass of solution = volume of solution × density = 100ml × 1.12 g/ml = 112g

now, mass of solvent = 112g - 40g = 72g

molality = number of mole of solute/mass of solvent in kg

= (40/58.5) × 1000/72

= 40000/(58.5 × 72)

≈ 9.5m

hence, 40% (w/v) NaCl solution is equivalent to 9.5m molality of solution.

Answer 2

Answer:

Molar mass of NaCl is 58.5

40% (W/V) NaCl solution

= 40 g Nacl in 100 ml solution

= 58.5 g Nacl in 100/40 * 58.5 ml solution

= 146.25 ml solution

specific gravity = 1.12

specific gravity = density of solute/ density of water

let density of solute=x

=>1.12=x/1

=>x= 1.12 g/ml

Density= Mass/Volume

=> Mass= Density* Volume

=> Mass= 1.12* 146.25

Mass= 163.8 g

Therefore ppm= mass of solute*10^6/ mass of solution

                        = 58.5*10^6/163.8

                        = 3.57*10^5 ppm