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Q.A sample of H2SO4H_2SO_4 is 94% (W/V)94\%\ (W/V) and its density is 1.841.84 g/mL. Calculate the molality of this solution. (H=1, O=16, S=32H = 1,\ O = 16,\ S = 32)

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2020Subjective· 2mImportance★★★★★
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From 9494 g acid in 184184 g solution: solvent =90=90 g, moles acid =0.959=0.959 ⇒ molality ≈10.66 m\approx10.66\,m.

Concept: Molality =moles of solutemass of solvent in kg=\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}. We must find the mass of water (solvent) in a fixed volume.

Step 1 — basis: 94%94\% (W/V) means 94 g H2SO4H_2SO_4 per 100 mL of solution.

Step 2 — mass of solution: =volume×density=100 mL×1.84 g mL−1=184=\text{volume}\times\text{density}=100\ \text{mL}\times1.84\ \text{g mL}^{-1}=184 g.

Step 3 — mass of solvent (water): =184−94=90 g=0.090=184-94=90\ \text{g}=0.090 kg.

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