Question

I was just wondering how to get these answers so when I’m faced with a problem like this again, I’ll know how to figure it out.

Answer 1

The question requires us to calculate the amount of a 0.250 M sucrose solution necessary to prepare 400.0 mL of a 0.0310 M solution.

The question provided the following information:

molarity of more concentrated solution = C(A) = 0.250 M = 0.250 mol/L

molarity of less concentrated solution = C(B) = 0.0310 M = 0.0310 mol/L

volume of less concentrated solution = V(B) = 400.0 mL

The question refers to a simple dilution process, where we use a more concentrated solution (A) to preare a less concentrated solution (B). The amount of sucrose in both solutions doesn't change, thus we can say that the number of moles of sucrose in A is the same as it is in B:

$n_A=n_B$

From the definition of molar concentration (also called molarity), we can say that the number of moles of a solution corresponds to its concentration (in mol/L) multiplied by its volume (in L):

$C=\frac{n}{V}\to n=C\times V$

Considering the two equations above, we can write:

$n_A=n_B\to C_A\times V_A=C_B\times V_B$

We can also rearrange this equation in order to calculate the volume of more concentrated solution, V(A), necessary:

$V_A=\frac{C_B\times V_B}{C_A}$

Now, we can use the values given by the question in the equation to calculate V(A):

$V_A=\frac{(0.0310\text{ mol/L)}\times(400.0\text{ mL)}}{(0.250\text{ mol/L)}}=49.6\text{ mL}$

Therefore, 49.6 mL of a 0.250 M sucrose solution are necessary to prepare 400.0 mL of a 0.0310 M solution. Note that, as 400.0 mL is the total volume of the solution B, this solution should be prepared with 49.6 mL of solution A + 350.4 mL of water.