Question

How does 0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution?

Answer 1

0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution has a mass ratio of 2: 1

Further explanation

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

• Molarity (M)

Molarity shows the number of moles of solute in every 1 liter of solution.

$\ large \ boxed {\ bold {M ~ = ~ \ frac {n} {V}}} [/ tex] Molality (m) Molality shows the number of moles dissolved in every 1000 grams of solvent. [tex]\large{\boxed{\boxed{\bold{m~=~n\times~\frac{1000}{p}}}}$

m = Molality

n = Number of moles of solute

p = Solvent mass (gram)

• Mol fraction (x)

The mole fraction shows the mole ratio of a substance to the mole of solution / mixture

$\large{\boxed{x~=~\frac{nA}{nA+nB}}}}$

nA + nB = 1

There are 0.5 m sucrose (molecular mass 342) solution and 0.5 m glucose (molecular mass 180)

Then the comparison of the two is from the ratio of mass (assuming the mass of the solvent is equal to 1 kg)

Sucrose (C₁₂H₂₂O₁₁) 0.5 molal then the mass:

mass = 0.5 m x 342 = 171 grams

Glucose (C₆H₁₂O₆) 0.5 molal then the mass:

mass = 0.5 m x 180 = 90 grams

So the mass ratio = 171: 90 = 1.9: 1 or we round it to 2: 1

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Keywords: molality, glucose, sucrose, mass, solvent,

Answer 2

Answer : Both solutions contain $3.011 X 10^{23}$ molecules.

Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain $3.011 x 10^{23}$ molecules.

Avogadro's Number is  $N_{A}$ =  $6.022 X 10^{23}$ which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.

Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules  = $N_{A} X M$;

∴  Number of molecules =  $6.022 X 10^{23} X 0.5 mol/L X 1 L$ which will be  = $3.011 X 10^{23}$

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.