Question

What is the molality of a solution which contains 36 g of glucose (C6H12O6) in 250 gm of water?

Answer 1

Answer : The molality of solution is, 0.8 mole/Kg

Solution : Given,

Mass of solute, glucose = 36 g

Mass of solvent, water = 250 g

Molar mass of glucose = 180 g/mole

Molality : It is defined as the number of moles of solute present in the one kilogram of solvent.

Formula used :

$Molality=\frac{w_b\times 1000}{M_b\times w_a}$

where,

$w_b$ = mass of solute, glucose

$w_a$ = mass of solvent, water

$M_b$ = molar mass of solute, glucose

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

$Molality=\frac{36g\times 1000}{180g/mole\times 250Kg}=0.8mole/Kg$

Therefore, the molality of solution is, 0.8 mole/Kg