Questions in solutions

SelectQuestion
A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. The freezing point of this solution is (Given ${{K}_{f}}={{1.86}^{o}}C/m$ for water) [IIT 1995]
${{K}_{f}}$ of 1, 4-dioxane is $4.9\ mo{{l}^{-1}}$ for $1000\ g$. The depression in freezing point for a $0.0001\ m$ solution in dioxane is
What is the freezing point of a solution containing 8.1g HBr in 100g water assuming the acid to be 90% ionised $({{K}_{f}}\,\text{for water}=1.86\,K\,mo{{l}^{-1}})$
0.440 g of a substance dissolved in 22.2 g of benzene lowered the freezing point of benzene by 0.567°. The molecular mass of the substance (${{K}_{b}}=5.12{}^\circ C\,mo{{l}^{-1}}$)
The molal freezing point constant for water is 1.86°C /mol. If 342 gm of cane sugar $({{C}_{12}}{{H}_{22}}{{O}_{11}})$ are dissolved in 1000 g of water, the solution will freeze at
What should be the freezing point of aqueous solution containing $17\ gm$ of ${{C}_{2}}{{H}_{5}}OH$ in $1000\ gm$ of water (water ${{K}_{f}}=1.86\ \deg -kg\ mo{{l}^{-1}}$)
A 0.001 molal solution of $\left[ Pt{{(N{{H}_{3}})}_{4}}C{{l}_{4}} \right]$in water had a freezing point depression of 0.0054°C. If ${{K}_{f}}$for water is 1.80, the correct formulation for the above molecule is
During depression of freezing point in a solution the following are in equilibrium
The freezing point of a solution prepared from 1.25 g of a non-electrolyte and 20 g of water is 271.9 K. If molar depression constant is $1.86\,K\,mo{{l}^{-1}}$ then molar mass of the solute will be
The freezing point of 1 molal $NaCl$ solution assuming $NaCl$ to be 100% dissociated in water is (molal depression constant = 1.86)

View Selected Questions (0)

Back to Categories

Back to Home