MHT CET Chemistry question

If 18 g of glucose is present in 1000 g of solvent, the solution is said to be

Answer: (D)

Solution

Moles of glucose = 18/180 = 0.1 mol in 1 kg of solvent, which is 0.1 mol per kg, i.e. 0.1 molal; "molar" refers

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Question ID VK-060906