After adding a solute freezing point of water solution decreases to ${- 0.186}^{o} C$ . Calculate $Δ T_{b}$ . ( $K_{f}$ $= {1.86}^{o} C k g m o l^{- 1}$ and $K_{b}$ $= {0.512}^{o} C k g m o l^{- 1}$ )

{0.521}^{o} C

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{0.0521}^{o} C

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{1.86}^{o} C

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{0.0186}^{o} C

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Solution

The correct option is B ${0.0521}^{o} C$
Given,
$Δ T_{f} = {0.186}^{0} C$

We know that,
$Δ T_{f} = K_{f} . m$

$\Rightarrow 0.186 = 1.86 \times m \Rightarrow m o l a l i t y = \frac{0.186}{1.86} = 0.1 m$
Using formula
$Δ T_{b} = K_{b} . m$

$= 0.521 \times 0.1 = 0.0521^{o} C$

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After adding a non volatile solute, freezing point of solution decreases to - 0.186. Calculate

After adding no – volatile solute freezing point of water decreases to $- {0.186}^{\circ} C$ . Calculate $Δ T_{b}$ if $K_{f}$ =1.86 K kg $m o l^{- 1}$ and $K_{b}$ = 0.521 K kg $m o l^{- 1}$

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After adding a solute freezing point of water solution decreases to −0.186oC. Calculate ΔTb. (Kf =1.86oCkgmol−1 and Kb =0.512oCkgmol−1)

The correct option is B 0.0521oCGiven,ΔTf=0.1860CWe know that,ΔTf=Kf.m⇒0.186=1.86×m⇒molality=0.1861.86=0.1 mUsing formulaΔTb=Kb.m=0.521×0.1=0.0521 oC

After adding a solute freezing point of water solution decreases to ${- 0.186}^{o} C$ . Calculate $Δ T_{b}$ . ( $K_{f}$ $= {1.86}^{o} C k g m o l^{- 1}$ and $K_{b}$ $= {0.512}^{o} C k g m o l^{- 1}$ )

The correct option is B ${0.0521}^{o} C$
Given,
$Δ T_{f} = {0.186}^{0} C$

We know that,
$Δ T_{f} = K_{f} . m$

$\Rightarrow 0.186 = 1.86 \times m \Rightarrow m o l a l i t y = \frac{0.186}{1.86} = 0.1 m$
Using formula
$Δ T_{b} = K_{b} . m$

$= 0.521 \times 0.1 = 0.0521^{o} C$