Question

if x% of a solute of molecular mass (Mo) is present in a solution having density of 'd' gm/cm3, calculate molarity,and molality,if
a. x% is by mass
b. x% is by volume​

Answer 1

The molality is $\frac{1000x}{M_0(100-x)} m$ and the molarity is $\frac{x*d*10}{M_0} M$ if x% is by mass.

And, the molality is $\frac{1000x}{M_0(100d-x)} m$ and the molarity is $\frac{10x}{M_0} M$ if the x% is by volume.

Given: Molecular mass of a solute = $M_o$

given % of solute = x %

the density of a solution = d gm/$cm^3$

To Find: Molarity and molality if

a) x% is by mass

b) x% is by volume.

Solution: a) x% by mass

It means we have x g in 100 g of solution

So, the mass of solvent = (100 - x) g

We know that number of moles of solute = $\frac{x}{M_0}$

Now, molality = $\frac{moles of solute}{mass of solvent(in kg)}$

= $\frac{1000x}{M_0(100-x)} m$

Given density = d

So, the volume of solution = $\frac{100}{d}$

Thus, molarity = $\frac{moles}{vol. of solution(in L)}$

= $\frac{x*d*1000}{M_0*100}$

= $\frac{x*d*10}{M_0} M$

So, the molality is $\frac{1000x}{M_0(100-x)} m$ and the molarity is $\frac{x*d*10}{M_0} M$ if x% is by mass.

b) x% is by volume

It means we have x g in 100 ml solution.

Molarity = $\frac{1000x}{100M_0}$

= $\frac{10x}{M_0} M$

Now, mass of solution = 100d

And, mass of solvent = (100d - x) g

So, molality = $\frac{moles of solute}{mass of solvent(in kg)}$

= $\frac{1000x}{M_0(100d-x)} m$

So, the molality is $\frac{1000x}{M_0(100d-x)} m$ and the molarity is $\frac{10x}{M_0} M$ if the x% is by volume.

#SPJ1