Question

Consider the following solutions,

(a) $4$ g of caustic soda is dissolved in $200$ mL of the solution.
(b) $5.3$ g of anhydrous sodium carbonate is dissolved in $100$ mL of solution.
(c) $0.365$ g of pure HCl gas is dissolved in $50$ mL of solution.

The molarity of these solutions respectively are :

• A. $0.5M,0.5M,0.5M$
• B. $0.2M,0.5M,0.2M$
• C. $0.5M,0.5M,0.2M$
• D. $0.2M,0.2M,0.5M$

Answer 1

The correct option is C $0.5M,0.5M,0.2M$

The expression for the molarity of the solution is, $Molarity=\frac{W}{M\times V}$
where,

$W$ represents weight of solute in g.
$M$ represents molar mass of solute in g/mol
$V$ represents the volume of the solution in L.
(a) The molarity of solution is$=\frac{4}{40\times 0.2}=0.5M.$
(b) The molarity of solution $=\frac{5.3}{106\times 0.1}=0.5M$.
(c) The molarity of solution $=\frac{0.365}{36.5\times 0.05}=0.2M$.