Match the List : [Atomic mass : Mg = 24, S = 32, Na = 23] List - I List - II (P) 6.3 gm HNO O 3 in 100 kg of solution (1) X solvent = 201 200 ( X = mole fraction ) (Q) 4 gm MgO and 360 gm water

Question

Match the List :

[Atomic mass : Mg = 24, S = 32, Na = 23]

List - I	List - II
(P) $6.3 gm HNO O_{3}$ in 100 kg of solution	(1) $X_{solvent} = \frac{200}{201} (X =$ mole fraction $)$
(Q) 4 gm MgO and 360 gm water	(2) $% \frac{W}{V} = 19.6$
(R) $2 M H_{2} SO_{4} (d_{solution} = 1.996 gm / mL)$	(3) 12 M
(S) $40% \frac{W}{W} N a O H$ solution $(d_{so l u t i o n} = 1.2$ gm/mL)	(4) 1 M
	(5) 63 ppm

TestHub · Accepted Answer

P: Mass of HNO₃ = 6.3 gm, Mass of solution = 100 kg = $100 \times 1 0^{3}$ gm.

ppm = $(\frac{ma sso f so l u t e}{ma sso f so l u t i o n}) \times 1 0^{6} = (\frac{6.3}{100 \times 1 0 ^{3}}) \times 1 0^{6} = 63$ ppm. So, P $\to$ 5.

Q: Moles of MgO = $\frac{4}{40} = 0.1$ mol. Moles of H₂O = $\frac{360}{18} = 20$ mol.

Mole fraction of solvent (H₂O) = $\frac{20}{20 + 0.1} = \frac{20}{20.1} \approx \frac{200}{201}$ . So, Q $\to$ 1.

R: Molarity (M) = 2 M H₂SO₄. Density of solution = 1.996 gm/mL.

Mass of 1 L solution = $1.996 \times 1000 = 1996$ gm.

Mass of H₂SO₄ = $2 \times 98 = 196$ gm.

Mass of solvent = $1996 - 196 = 1800$ gm = 1.8 kg.

Molality (m) = $\frac{2 m o l}{1.8 k g} = 1.11$ m.

% W/V = $(\frac{ma sso f so l u t e}{v o l u m eo f so l u t i o n}) \times 100 = (\frac{196 g m}{1000 m L}) \times 100 = 19.6%$ . So, R $\to$ 2.

S: 40% w/w NaOH solution. Density = 1.2 gm/mL.

Consider 100 gm solution. Mass of NaOH = 40 gm. Mass of solvent = 60 gm.

Volume of solution = $\frac{100 g m}{1.2 g m / m L} = 83.33$ mL.

Molarity = $\frac{m o l eso f N a O H}{v o l u m eo f so l u t i o nin L} = \frac{40/40}{83.33/1000} = \frac{1}{0.08333} = 12$ M. So, S $\to$ 3.

Therefore, P $\to$ 5; Q $\to$ 1; R $\to$ 2; S $\to$ 3.

Chemistry - Concentration Terms Question with Solution | TestHub

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JEE Advanced 2027 (Class 12)