Chapter U4.81, Problem 5E

Solution

(a)

Interpretation:

The solutioncontaining most molecule must be explained among 1.0 L of 1.0 M glucose, 1.0 L of 1.0 M sucrose and 500 ml of 1.0 M sucrose.

Concept Introduction :

Moles of solute can be found from the following formula.

  $n=M\times V$

Here, M is molarity and V is volume of the solution in L.

The number of molecules can be calculated from moles and Avogadro’s number as follows:

  $N=n\times N_A$

Both 1.0 L of 1.0 M glucose and 1.0 L of 1.0 M sucrose have the most molecules.

Moles of glucose in 1.0 L of 1.0 M glucose is $1.0\times 1.0=1$ .

Number of molecules= $1\times N_A$ .

Moles of sucrose in 1.0 L of 1.0 M sucrose is $1.0\times 1.0=1$ .

Number of molecules= $1\times N_A$ .

Moles of sucrose in 500 mL of 1.0 M sucrose is $1.0\times 0.5=0.5$ .

Number of molecules= $0.5\times N_A$ .

Thus, both 1.0 L of 1.0 M glucose and 1.0 L of 1.0 M sucrose have the same molecules and the number is double the number of molecules in 500 mL of 1.0 M sucrose.

(b)

Interpretation:

The solution with thegreatest concentration must be explained among 1.0 L of 1.0 M glucose, 1.0 L of 1.0 M sucrose and 500 ml of 1.0 M sucrose.

Concept Introduction :

Molarity is the number of moles of solute dissolved in 1 L solution.

It is one concentration unit of solution.

All the three solutions have same concentration.

Molarity of three solutions is 1.0 M.

Volumes of two solutions are 1.0 L. One sucrose solution has volume 500 mL or 0.5 L.

So a concentration of all the three solutions is same.

(c)

Interpretation:

The solution which has most mass must be explained among 1.0 L of 1.0 M glucose, 1.0 L of 1.0 M sucrose and 500 ml of 1.0 M sucrose.

Concept Introduction :

Mass of solution is the total mass of solute and solvent.

Mass of solute can be calculated from number of moles and molar mass as follows:

  $m=n\times M$

1. L of 1.0 M sucrosehas the most mass.

Moles of glucose in 1.0 L of 1.0 M glucose is 1 mol

Mass of glucose (molecular weight 180) is $1\times 180g=180g$ .

Now,

Moles of sucrose in 1.0 L of 1.0 M sucrose is 1 mol.

Mass of sucrose (molecular weight 342) is $1\times 342g=342g$ .

Similarly,

Moles of sucrose in 500 mL of 1.0 m sucrose is 0.5.

Mass of sucrose is $0.5\times 342g=171g$ .