Question

1.04gram of bacl2 is present in 100000 gram solution than what is the concentration of solution is

Answer 1

Molality = moles /mass of solvent in kg

= 1.04/208×1000/100000

= 5×10 raise to -5 molal.

Answer 2

Answer: $5\times 10^{-5}mole/kg$

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n}{W_s}$

where,

n = moles of solute

 $W_s$ = weight of solvent in kg

moles of solute =$\frac{\text {given mass}}{\text {molar mass}}=\frac{1.04g}{208g/mol}=0.005moles$

mass of solution = 100000 g

mass of solute = 1.04 g

mass of solvent = mass of solution - mass of solute = (100000- 1.04) g = 99998.96 g = 99.9 kg

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.005moles}{99.9kg}=5\times 10^{-5}mole/kg$

Therefore, the molality of solution will be $5\times 10^{-5}mole/kg$