Question

0.01 mole sucrose (C_(12)H_(22)O_(11) ) dissolve in ......... litre water so it becomes 0.01m solution.

Answer 1

Answer:

Can you prepare 1 liter of a 0.5-molar glucose solution? ... 0.5-molar glucose = 90 g. Add this mass of the compound to water bring it to volume of 1 liter. This makes 1 liter of 0.5 M solution.

Answer 2

Answer:

0.01 mole sucrose C₁₂H₂₂O₁₁ dissolve in one liter of water to get solution of 0.01m.

Explanation:

Given, the number of moles of C₁₂H₂₂O₁₁ = 0.01moles

The molecular mass of sucrose = 12(12)+22(1)+11(16) = 342gmol⁻¹

Mass of sucrose dissolved = Molar mass × moles

                                           = 342×0.01 =3.42g

Molality of a solution can be calculated by dividing the number of moles of solute to mass of solvent in kilograms.

We have given the molality of sucrose solution = 0.01m

Molality of a solution = (Moles of solute)/(Mass of solvent in Kg)

$0.01=\frac{0.01}{Mass \;\; of\;\; solvent}$

Mass of water(solvent) = 0.01/0.01= 1kg

We know, density of water = 1gml⁻¹

Volume of water = mass/density

Volume of water $=\frac{10^{3}g}{1gml^{-1} }=1000ml =1L$

Therefore, 0.01 mole of sucrose dissolve in 1L of water.