Question

calculate the molality of solution containing 3 g of glucose dissolved in 30 g of water

Answer 1

Molality is known as moles of solute dissolved in per kg of solvent
So 1st we need to use molar mass of glucose to find moles in sample
3 * 1 mole of glucose / 180.16 g = 0.01665 moles og glucose
We know the moles of glucose disolved in 30 g 0f water
Now we need to sum up for per kg of water so converting mass of water from gram to kg
30g * 1kg / 10^3 g = 0.030 kg
Now 0.030 kg water contain 0.01665 moles of glucose . I kg of water will contain:
1 kg water * 0.01665 moles of glucose / 0.030 kg of water = 0.555 moles of glucose
So molality of solution is 0.6 molal solution

Answer 2

In order to know your answer firstly know about molality
Molality may be defined as the number of moles of solute that are  dissloved in a kg of water .
so firstly we have to calculate the moles of glucose
3*1/180.16 = 0.01665 ( as the molar mass of glucose= 180.1559)
These are dissloved in 30 g of water.
first convert mass into kilogram =30/1000= 0.030 kg
0.030 kg water = 0.01665 moles of glucose
so 1 kg of water=1*0.01665/0.030 = 0.555 moles of glucose
round about of 0.555 will be 0.6 which is our answer