Question

What is the (approx) molarity of 1.5 molal of H2SO3 solution (density of solution=0.75g/cc)

Answer 1

Answer : The molarity of solution is, 1.01 mole/L

Explanation :

The relation between the molarity, molality and the density of the solution is,

$d=M[\frac{1}{m}+\frac{M_b}{1000}]$

where,

d = density of solution  = $0.75g/cm^3=0.75g/ml$

m = molality of solution  = 1.5 mole/Kg

M = molarity of solution  = ?

$M_b$ = molar mass of solute $(H_2SO_3)$ = 82 g/mole

Now put all the given values in the above formula, we get

$0.75g/ml=M[\frac{1}{1.5mole/Kg}+\frac{82g/mole}{1000}]$

$M=1.01mole/L$

Therefore, the molarity of solution is, 1.01 mole/L

Answer 2

Answer:

The molarity of the solution is 1 M.

Hope it will help

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