Calculate the cryoscopic constant $(K_f)$ of a solvent if the depression in freezing point of a $0.18 \ m$ solution of a non-volatile solute is $0.2 \ K$.
- A$1.73 \ K \ kg \ mol^{-1}$
- B$1.50 \ K \ kg \ mol^{-1}$
- C$1.11 \ K \ kg \ mol^{-1}$
- D$1.92 \ K \ kg \ mol^{-1}$
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$3 \times 10^{-3} \ kg$ acetic acid is added into $500 \ cm^{3}$ water. If dissociation of acetic acid is $23\%$ then find out depression in freezing point? $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$ and density $= 0.997 \ g \ cm^{-3}$.
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View SolutionAn aqueous solution of a weak monobasic acid containing $0.1 \text{ g}$ in $21.7 \text{ g}$ of water freezes at $272.813 \text{ K}$. If the value of $K_f$ for water is $1.86 \text{ K kg/mol}$,what is the molecular mass of the monobasic acid in $\text{g/mol}$?
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View SolutionWhich of the following formulae is used to calculate the depression in freezing point?
Calculate the cryoscopic constant $(K_f)$ when $0.8 \ g$ of a non-volatile solute with a molar mass of $64 \ g \ mol^{-1}$ is dissolved in $43 \ g$ of a solvent,lowering the freezing point by $0.34 \ K$.
What is the depression of freezing point,when mole fraction of non-electrolyte solute in aqueous solution is $0.01$ (in $K$)? ($K_f$ of $H_2O = 1.86 \ K \ kg \ mol^{-1}$)
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