Question

Calculate the freezing point of a solution when 3 gram of cacl2 was dissolved in hundred gram of water assuming cacl2 undergoes complete ionization

Answer 1

Answer: Del Tf = i Kf molality

Explanation: = 3x1.86 x3/111 *1000/100

To - Ts = 1.508

To =freezingf point of solvent

Ts = freezing point of solution

0 - Ts = 1.508

Ts = - 1.508

Answer 2

The freezing point of a solution when 3 gram of $CaCl_2$ was dissolved in hundred gram of water assuming $CaCl_2$ undergoes complete ionization is $-1.51^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-T_f)^0C$ = Depression in freezing point

i= vant hoff factor for $CaCl_2$, i= 3 as it is a electrolyte and dissociate to give 3 ions on complete  ionization

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

$K_f$ = freezing point constant for water= $1.86^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (water)= 100 g = 0.1 kg     (1kg=1000g)

Molar mass of solute $(CaCl_2)$ = 111 g/mol

Mass of  solute $(CaCl_2)$ added = 3 g

$(0-T_f)^0C=3\times 1.86\times \frac{3g}{111 g/mol\times 0.1kg}$

$(0-T_f)^0C=1.51$

$T_f=-1.51^0C$

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