The correct option is D P−3,Q−2,R−1,S−4
(P)
4.5 m solution means 4.5 mol of CaCO3 is present in 1000 g of the solvent.
So, mass of the solute
= given moles× molar mass
= 4.5× 100 = 450 g
Total mass of the solution
= 450 g + 1000 g = 1450 g
Now, density = mass of the solutionvolume of the solution
Volume of the solution
= 14501.45= 1000 mL
Molarity = moles of the solutevolume of the solution in L
= 4.51 = 4.5 M
(Q)
We know, M1V1+M2V2=M3V3
where, M1 and V1 =3 M and 100 mL HCl
& M2 and V2 = 1 M and 300 mL HCl
Total volume after mixing , V3 = 400 mL
So, M = 3×100+1×300400
=1.5 M of HCl
(R)
14.5 m solution of Ca2+ solution means 14.5 moles are present in 1 kg of the solvent (Water).
moles of solvent = 100018 = 55.56
mole fraction of Ca2+
=14.514.5+55.56 = 0.2
(S)
Molar mass of NaOH = 40 g/mol
% w/w = mass of NaOHtotal mass of solution×100
=40×55×40+1000×100 = 16.66%