Question

calculate the molality of solution containing 3 gram glucose dissolve in 30 gram of water ​

Answer 1

Given:-

• 3 gram glucose dissolve in 30 gram of water.

To Find:-

• The Molality.

Solution:-

We Know that

Molality- Molality is number of moles of solute dissolved in a certain mass of solvent. It is defined as the moles of a solute per kilograms of a solvent.

So, Formula of Molality:-

$Molality = \frac{no. \: of \: moles}{mass \: of \: solvent}$

And

$\sf \: Number \: of \: moles = \frac{given \: mass}{molecular \: mass}$

$given \: mass = 3g$

$molecular \: mass = 180g$

So,

$= > Number \: of \: moles = \frac{3}{180}$

$= > 0.0166$

After this it is given that these are dissolved in 30g of water.

Now We need to convert 30g into kg

$= > \frac{30}{1000} = 0.03kg$

So, According to the Question:

0.03kg of water=0.0166 moles of glucose

So,For 1kg of water:

= >1 kg of water= $\frac{0.0166}{0.03}$ moles of glucose

$= > 0.55$

∴ the molality of solution containing 3 gram glucose dissolve in 30 gram of water is 0.55

Answer 2

Explanation:

Mass of glucose = 2.82 g 

No. of moles of glucose = 2.821802.82180= 0.016

Mass of water = 30g = 30183018 = 1.67

hope it will help you