Question

What is the mass percent and molarity of water/sucrose in 0.302g of H2O and 0.014g of C12H22O11?

Answer 1

The mass % can be calculated y using equation,

Mass % = ( Mass of substance / Total mass of compound ) x 100

The total mass of compound = 0.302 g water + 0.014 g C₁₂H₂₂O₁₁

  = 0.316 g

1) Mass % of water = ( 0.302 / 0.316 ) x 100

= 0.9556 x 100

= 95.56 %

2) Mass % of  C₁₂H₂₂O₁₁ = ( 0.014 / 0.316 ) x 100

= 0.0443 x100

= 4.43 %

3) Molarity can be calculate as. Molarity = Moles / Total volume in lit

a ) Moles of water = Mass / Molar mass

= 0.302 g / 18 g/mole

= 0.0167 moles

b) Moles of Sucrose = Mass / Molar mass

= 0.014 / 342.3

= 0.00004089 moles

The density of water is 1 so volume of water is 1 x 0.302 g = 0.302 mL

= 0.302 / 1000 = 0.000302 Lit

So molarity of Sucrose = Moles / Volume

= 0.00004089 / 0.000302

= 0.13 M