Question

5% glucose solution (M (C 6 H 12 O 6 ) = 180 g/mol) is used to replenish fluid in the body and as a source
of easily digestible carbohydrate. Calculate molarity and molality of this solution (ρ solution =1.02 g/ml).

Answer 1

The molarity and the  molality of the solution are 0.29 M and 0.295m.

Given:

Percentage of glucose in the solution = 5%

The solution has a density of 1.02 g/ml.

M($C_{6} H_{12} O_{6}$) =180 g/mol

To find:

The molarity and the molality of the given solution.

Solution:

Step 1 :Calculating the no:of moles.

5% glucose solution means that in the solution has 5g of glucose(Solute) and 95g of  water(Solvent)

$the number of moles in the solute = \frac{mass of the solute}{Molar mass of t he solute}$

$= \frac{5}{180}$

$= 0.028$.

Step 2: Calculating the molarity of the solution.

It's given that density 1.02 g/ml

So,

Volume of the solution = $\frac{1}{10.2} litres$  [ by conversion $\frac{100}{1.02} * \frac{1}{1000}$ = $\frac{1}{10.2} L$ ]

Now

$Molarity of the solution = \frac{No:of moles of the solute}{Volume of the solution in litres}$

$= \frac{0.028}{\frac{1}{10.2} }$

$= 0.028 * 10.2$

$= 0.29M$

Step 3: Calculating the Molality of the solution.

Then Molality of the solution is given by $\frac{No: of moles }{Mass of solvent in Kg}$

Substituting the corresponding values we get,

$Molality of the solution = \frac{0.028}{\frac{95}{1000} }$

$= \frac{0.028 * 1000}{95}$

$= 0.295 m$

Therefore, the Molarity of the solution is 0.29M and the Molality of the solution is 0.295

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