After adding no - volatile solute freezing point of water decreases to -0-186-C- Calculate - Tb if Kf-1-86 K kg mol-1 and Kb - 0-521 K kg mol-1

∴ T_f =K_f × m ∴ m = 0.1 0.186 = 1.86 × m ∴ m =0.1 Δ T_b = K_b × m ∴ 0.521 × 0.1 = 0.0521 ...

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Standard XII

Chemistry

Abnormal Colligative Properties

After adding ...

Question

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 $m o l^{- 1}$

0.521

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0.0521

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1.86

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0.0186

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Solution

The correct option is B
0.0521

$∴ T_{f} = K_{f} \times m ∴ m = 0.1$
$0.186 = 1.86 \times m ∴ m = 0.1$
$Δ T_{b} = K_{b} \times m ∴ 0.521 \times 0.1 = 0.0521$

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Similar questions

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 no – volatile solute freezing point of water decreases to −0.186∘C. Calculate ΔTb if Kf=1.86 K kg mol−1 and Kb = 0.521 K kg mol−1

The correct option is B 0.0521 ∴Tf=Kf×m∴m=0.1 0.186=1.86×m∴m=0.1 ΔTb=Kb×m∴0.521×0.1=0.0521

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 $m o l^{- 1}$

The correct option is B
0.0521

$∴ T_{f} = K_{f} \times m ∴ m = 0.1$
$0.186 = 1.86 \times m ∴ m = 0.1$
$Δ T_{b} = K_{b} \times m ∴ 0.521 \times 0.1 = 0.0521$