Question

1. 4 g of glucose C6H12O6 were dissolved in 50 g of water. Calculate molarity of the solution and percent by mass, if density r of the solution is 1.379 g/mL​

Answer 1

The molarity is 0.06M and mass % is 7.4%

GIVEN

4 g of glucose C₆H₁₂O₆ were dissolved in 50 g of water. Density of the solution is 1.379 g/mL.

TO FIND

The molarity of the solution and percent by mass.

SOLUTION

We can simply solve the above problem as follows;

Mass of solute (C₆H₁₂O₆) = 4 grams

Mass of Solvent = 50 grams

Mass of solution = 50 + 4 = 54 grams

Density of solution = 1.379 g/ml

We know that,

Density = Mass/Volume

1.39 = 54/Volume

Volume = 45/1.39 = 32.37 ml

We know that,

Molarity = Moles of solute/volume of solution in litres

Moles of solute = 4/180 = 0.02

Molarity = (0.02 × 100)/32.27

= 0.06 M

Mass% = (Mass of solute/Mass of solution)× 100

= (4/54)× 100

= 7.4%

Hence, The molarity is 0.06M and mass % is 7.4%

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