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Q.A solution containing 10.2 g of glycerine per litre is found to be isotonic with 2% solution of glucose. Calculate the molar mass of glycerine. (Molar mass of glucose = 180 g mol⁻¹) OR A solution of sucrose (Molar mass = 342 g mol⁻¹) is prepared by dissolving 70 g of it per litre of solution. What is the osmotic pressure at 27°C ? (R = 0.082 L bar K⁻¹ mol⁻¹)

Mizoram MbseMizoram Board of School Education HSSLC 2025Subjective· 4mImportance★★★★★
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Isotonic solutions have equal molar concentration; equating the molarity of the 2% glucose solution to the molarity implied by 10.2 g/L of glycerine gives a molar mass of about 91.8 g/mol.

Step 1 — Molarity of the 2% glucose solution:

2% (w/v) glucose means 2 g of glucose per 100 mL of solution, i.e. 20 g per litre.

Molarity of glucose=20 g/L180 g/mol=0.1111 mol/L\text{Molarity of glucose} = \dfrac{20\ \text{g/L}}{180\ \text{g/mol}} = 0.1111\ \text{mol/L}

Step 2 — Isotonic condition:

Two solutions are isotonic when they exert the same osmotic pressure at the same temperature, which (for non-electrolytes, same T) means they must have the same molar concentration:

Cglycerine=Cglucose=0.1111 mol/LC_{\text{glycerine}} = C_{\text{glucose}} = 0.1111\ \text{mol/L}

Step 3 — Molar mass of glycerine:

Given concentration of glycerine = 10.2 g/L. Using C=massMolar mass×VC = \dfrac{\text{mass}}{\text{Molar mass} \times V}: …

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