Question

The molarity of a glucose containing 36 g of glucose per 400 ml of the solution is:
A.1.0
B.0.5
C.2.0
D.0.05

Answer 1

Hint:We can express the concentration of a solution by different methods. Which include Molarity, Normality, Molality, Mole fraction, Mass fraction, Mass fraction, Mass percentage, Volume percentage.


Complete step by step answer:
Molarity is defined as the number of moles of a solute present per liter of the solution.
Molarity= Number of moles of the solute/ Volume of solution in liters.
 $c = \dfrac{n}{v}$
Where n is the number of moles of solute and V is the volume of the solution in Liters.
If the number of moles of a solute present in V liter of its solution is n2, the molarity of the solution is given by
 $M = \dfrac{{n2}}{V}$
Calculation:
  \[No.\,of\,Moles\,of\,solute = \dfrac{{Mass\,in\,g}}{{Gram\,molecular\,mass}}\]
  \[No.\,of\,Moles\,of\,Glucose,n = \dfrac{{36}}{{180}}\] =0.2 moles
 \[No.\,of\,moles\,of\,a\,gas = \dfrac{{V0}}{{22.414}}\] V Is the volume of solution in Liters. We are taking 400 ml of solution.
The volume of solution in liter, $V = \dfrac{{400}}{{1000}} = 0.4$
 \[Molarity\,is\,given\,by\,c = \dfrac{n}{v}\]
 $c = \dfrac{{0.2}}{{0.4}}$
C=0.5 M
Hence, the correct answer is option (B) i.e 0.5 M

Note: We can find the number of molecules of the sample,
 \[Number\,of\,Molecule = Number\,of\,moles \times \,Avogadro\,Number\]
180 g of Glucose contains \[6.022 \times {10^{23}}\] glucose molecules. So, 36 g of Glucose contains
 \[\dfrac{{6.022 \times {{10}^{23}}}}{{180}} \times \left[ {36} \right]\] number of molecules.
If we are given with a gas, then \[No.\,of\,moles\,of\,a\,gas = \dfrac{{V0}}{{22.414}}\]