Question

Match the elements given in List - I with their correct option given in List - II.

$\begin{array}{ccc}& \text{List-I}& \text{List - II}\\ \text{P.}& \text{4.5 m solution of}CaC{O}_3\text{of density 1.45 g/mL}& 1.\text{mole fraction of solute is 0.2}\\ \text{Q.}& \text{3 M 100 mL}HCl\text{mixed with 1 M 300 mL solution of}HCl& 2.\text{Molarity = 1.5 M}\\ \text{R.}& \text{14.5 m solution of}C{a}^{2+}& 3.\text{Molarity = 4.5 M}\\ \text{S.}& \text{5 m}NaOH\text{solution}& 4.\text{16.66 \% w/w of}NaOH\text{in the solution}\end{array}$

• A. $P-4,Q-2,R-3,S-1$
• B. $P-2,Q-3,R-1,S-4$
• C. $P-2,Q-1,R-3,S-4$
• D. $P-3,Q-2,R-1,S-4$

Answer 1

The correct option is D $P-3,Q-2,R-1,S-4$

(P)
4.5 m solution means 4.5 mol of $CaC{O}_3$ is present in 1000 g of the solvent.
So, mass of the solute
= given moles$\times$ molar mass
= 4.5$\times$ 100 = 450 g
Total mass of the solution
= 450 g + 1000 g = 1450 g

Now, density = $\frac{\text{mass of the solution}}{\text{volume of the solution}}$

Volume of the solution
= $\frac{1450}{1.45}=$ 1000 mL

Molarity = $\frac{\text{moles of the solute}}{\text{volume of the solution in L}}$
= $\frac{4.5}{1}$ = 4.5 M

(Q)
We know, $M_1{V}_1+M_2{V}_2=M_3{V}_3$
where, $M_{1}and{V}_1=3M\text{and}100mLHCl$
& $M_{2}and{V}_2$ = 1 M and 300 mL $HCl$

Total volume after mixing , $V_3$ = 400 mL
So, M = $\frac{3\times 100+1\times 300}{400}$
=1.5 M of HCl

(R)
14.5 m solution of $C{a}^{2+}$ solution means 14.5 moles are present in 1 kg of the solvent (Water).
moles of solvent = $\frac{1000}{18}$ = 55.56
mole fraction of $C{a}^{2+}$
$=\frac{14.5}{14.5+55.56}$ = 0.2

(S)
Molar mass of $NaOH$ = 40 g/mol

% w/w = $\frac{\text{mass of NaOH}}{\text{total mass of solution}}\times 100$

$=\frac{40\times 5}{5\times 40+1000}\times 100$ = 16.66%