Question

The following successive dilutions are applied to a stock solution of 5 M sucrose (molecular weight = 342):

Solution A: 50 mL of stock solution is diluted to 250 mL
Solution B: 50 mL of solution A is diluted to 100 mL
Solution C: 100 mL of solution B is diluted to 1000 mL

What is the concentration, in %(w/V) of sucrose in solution C?

• A. 3.42%
• B. 1.71%
• C. 17.1%
• D. 34.2%

Answer 1

The correct option is B 1.71%

Initial concentration $\left(M_o\right)$ for sucrose = 5 M
Applying mole conservation for solution A,
$M_o\times V_o=M_A\times V_A$,
where, $M_{A}and{V}_A$ are the concentration and the volume of solution A respectively.
$\to M_A=\frac{5\times 50}{250}=1M$
Similarly, for solution B,
$M_A\times V_A^{{}^{\prime }}=M_B\times V_B$
where, $M_{B}and{V}_B$ are the concentration and the volume of solution B respectively.
$\to M_B=\frac{1\times 50}{100}=0.5M$
For solution C,
$M_B\times V_B^{{}^{\prime }}=M_C\times V_C$
where, $M_{C}and{V}_C$ are the concentration and the volume of solution C respectively.
$\to M_C=\frac{0.5\times 100}{1000}=0.05M$
Now, 0.05 mol is present in 1 L of solution C.

Mass of sucrose in solution C
= 0.05$\times$342 = 17.1 g
Concentration in %(w/V) = $\frac{17.1}{1000}\times 100=1.71$%