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Q.A 1% solution of solute 'X' is isotonic with a 6% solution of sucrose (molar mass = 342 g mol−1mol^{-1}). The molar mass of solute 'X' is : (A) 34·2 g mol−1mol^{-1} (B) 57 g mol−1mol^{-1} (C) 114 g mol−1mol^{-1} (D) 3·42 g mol−1mol^{-1}

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Isotonic solutions have the same osmotic pressure, which for dilute non‑electrolytes means equal molar concentrations. Equating the molarities of the 1% X solution and the 6% sucrose solution gives the molar mass of X as 57 g mol⁻¹.

The key idea here is that isotonic solutions exert the same osmotic pressure. For dilute solutions of non‑electrolytes (like sucrose and the unknown solute X), osmotic pressure is given by Π=iCRT\Pi = iCRT, and since neither solute dissociates, i=1i = 1. So Π\Pi depends only on the molar concentration CC (in mol L⁻¹) at a given temperature. If two solutions are isotonic, their molar concentrations must be equal.

The problem gives us percentage concentrations — 1% of X and 6% of sucrose. A “1% solution” means 1 g of solute in 100 mL of solution (or equivalently 10 g per litre). Similarly, 6% sucrose means 6 g per 100 mL, i.e. 60 g per litre. We can convert these mass‑per‑volume concentrations into molarities using the molar mass, and then set them equal.

Let’s work through it step by step.

  1. Write the expression for molarity of each solution.

    Molarity M=mass of solute per litre (g L⁻¹)molar mass (g mol⁻¹)M = \frac{\text{mass of solute per litre (g L⁻¹)}}{\text{molar mass (g mol⁻¹)}}.

    For sucrose: Msucrose=60342M_{\text{sucrose}} = \frac{60}{342} mol L⁻¹.

    For X: MX=10MXM_X = \frac{10}{M_X} mol L⁻¹, where MXM_X is the unknown molar mass in g mol⁻¹.

  2. Set the molarities equal because the solutions are isotonic.

10MX=60342\frac{10}{M_X} = \frac{60}{342}

  1. Solve for MXM_X. Cross‑multiply: 10×342=60×MX10 \times 342 = 60 \times M_X

3420=60 MX3420 = 60\,M_X

MX=342060=57M_X = \frac{3420}{60} = 57 …

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