Question

29.2% (w/w) HCl stock solution has a density of 1.25 $g\quad ml^ -1$. The molecular weight of HCl is 36.5 $g\quad mol^ -1$. The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is:
A) 8ml
B) 7ml
C) 6ml
D) 5ml

Answer 1

From w/w % find out the weight of HCl per 100 g of solution. Calculate the volume of 100 g of solution and then calculate the molarity of the stock solution. Find out the volume of stock solution required by equating the number of moles on both sides.

Complete answer:
We first find out the weight of HCl per 100 g of solution.
Given w/w% is 29.2 it means that 100 g of solution contains 29.2 g of HCl. Let us consider a 100 g of solution and calculate the molarity of the solution. The weight of HCl will be 29.2 g in 100 g solution.
Molarity
$M\quad =\quad \dfrac { Number\;of\; moles\; of\; solute }{ Volume\; of\; solution\; in\; litres}$

Number of moles of HCl is $\dfrac { 29.2 }{ 36.5 }$.

The volume of solution in litres is $\dfrac { Weight\; of\; solution }{ Density\; of\; solution } =\dfrac { 100g }{ 1.25g\quad cm^{ -3 } } = 80\;cm^{ 3 } = 80ml$

$M =\dfrac { 0.8 }{ 80 } \times 1000 = 10M$

Molarity of stock solution is 10M.
We know that a few ml of the stock is diluted to form 200 mL of 0.4 M HCl. Therefore, the number of moles HCl in-stock solution will be equal to the number of moles in a diluted solution.
$M_{ 1 }{ V }_{ 1 }\quad =\quad { M }_{ 2 }{ V }_{ 2 }$
$M_{ 1 },\quad M_{ 2 }$ are molarities of first and second solutions respectively.
$V_{ 1 },\quad V_{ 2 }$ are volumes of first and second solutions respectively.
Molarity of first solution = 10M, molarity of second solution = 0.4M
Let volume of first solution be V, volume of second solution = 200ml
$10\quad \times \quad V\quad =\quad 0.4\quad \times \quad 200$

$V\quad =\quad \dfrac { 4\quad \times \quad 20 }{ 10 } \quad =\quad \dfrac { 80 }{ 10 } \quad =\quad 8ml$

Therefore, the volume of the stock solution required is 8ml. Hence option A is correct.

Note: While calculating molarity we need to take the volume of the entire solution and should not consider only the volume of solvent. We need to be careful with units and take volume in litres while calculating molarity.