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Arrange the following aqueous solutions in the order of their increasing boiling points (i) ${{10}^{-4}}M\,NaCl$ (ii) ${{10}^{-4}}M$urea (iii) ${{10}^{-3}}M\,MgC{{l}_{2}}$ (iv) ${{10}^{-2}}M\,NaCl$
10 g of solute with molecular mass 100 g $mo{{l}^{-1}}$is dissolved in 100 g of solvent to show 0.3° elevation in boiling point. The value of molal ebullioscopic constant will be
The boiling point of 0.1 molal ${{K}_{4}}\left[ Fe{{(CN)}_{6}} \right]$solution will be (given ${{K}_{b}}$for water $=0.52{}^\circ \text{C}\,\text{kg}\,\text{mo}{{\text{l}}^{\text{-1}}}$)
If the elevation in boiling point of a solution of 10 gm of solute (mol. wt. =100) in 100 g of water is $\Delta {{T}_{b}}$ , the ebullioscopic constant of water is
The freezing point of a solution containing 4.8 g of a compound in 60 g of benzene is 4.48. What is the molar mass of the compound $({{K}_{f}}=5.1\,K\,{{m}^{-1}},$freezing point benzene $={{5.5}^{o}}C$)
An aqueous solution freezes at $-{{0.186}^{o}}C$ (${{k}_{f}}={{1.86}^{o}};\ {{k}_{b}}={{0.512}^{o}}$). What is the elevation in boiling point
Molal depression constant for water is ${{1.86}^{o}}C$. The freezing point of a 0.05 molal solution of a non-electrolyte in water is
The freezing point of a $0.01\ M$ aqueous glucose solution at 1 atmosphere is $-{{0.18}^{o}}C$. To it, an addition of equal volume of $0.002\ M$ glucose solution will produce a solution with freezing point of nearly
When 0.01 mole of sugar is dissolved in $100g$ of a solvent, the depression in freezing point is ${{0.40}^{o}}$. When 0.03 mole of glucose is dissolved in $50g$ of the same solvent, the depression in freezing point will be
The freezing point of 0.05 molal solution of a non-electrolyte in water is

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