Calculate the molality of a solution containing a non-volatile solute if the solution freezes at $-0.95^{\circ}C$ $[K_f \text{ for water} = 1.86 \ K \ kg \ mol^{-1}, \text{ freezing point of water} = 0^{\circ}C]$.
- A$0.51 \ mol \ kg^{-1}$
- B$0.41 \ mol \ kg^{-1}$
- C$0.75 \ mol \ kg^{-1}$
- D$0.65 \ mol \ kg^{-1}$
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When $1.25 \ g$ of a non-volatile solute is dissolved in $20 \ g$ of water,the freezing point of the solution is found to be $271.9 \ K$. If the molal depression constant $(K_f)$ is $1.86 \ K \ kg \ mol^{-1}$,what is the molar mass of the solute?
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View SolutionAn aqueous solution containing $0.2 \ g$ of a non-volatile solute '$A$' in $21.5 \ g$ of water freezes at $272.814 \ K$. If the freezing point of water is $273.16 \ K$,the molar mass (in $g \ mol^{-1}$) of solute '$A$' is $[K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}]$
What is the molar mass of a solute when $5 \ g$ of solute dissolved in $70 \ g$ of solvent lowers its freezing point by $2.5 \ K$? Given $K_f = 3.5 \ K \ kg \ mol^{-1}$.
$A$ solution of $5 \ g$ nonvolatile solute in $50 \ g$ water decreases its freezing point by $0.2 \ K$. Calculate the molar mass of solute if $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$.
$2 \ g$ of a non-electrolyte solute (molar mass is $500 \ g \ mol^{-1}$) was dissolved in $57.3 \ g$ of xylene. If the freezing point depression constant $K_f$ of xylene is $4.3 \ K \ kg \ mol^{-1}$,then the depression in freezing point of xylene is.......... (in $K$)
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