Question

the molality of 0.05 M HCL(aq.) in 5L solution at 4 degree C is
options
0.25m
0.05m
5.0m
2.5m

Answer 1

Answer:

correct answer is 0.05m

Explanation:

volume of solution 5 litre = 5000ml

molarity of HCL(aq)  is 0.05 M

molarity = no of moles of HCl / volume of solution

no of moles of Hcl = molarity * volume of solution

no of moles of HCl = 0.05 * 5

no of moles of HCl =  0.25 mole

mass of HCL = moles * molar mass

molar mass of HCl = 36.5

mass of HCl = 0.25 * 36.5

mass of HCl = 9.125 gm

at 4 degree celsius  density of water is 1 gm/cm^3

mass of solution =  density * volume of solution in ml

mass of solution = 1 gm/cm^3  *5000 ml

mass of solution  = 5000gm

here we can see

mass of solution >>>>>> mass of solute (HCl)

therefore we can say,

mass of solution = mass of solvent = 5000gm = 5 Kg

molality = moles of solute(HCl) / mass of solvent in Kg

molality  = 0.25 mole/5 Kg

molality  =  0.05m

molality of solution is 0.05m