An aqueous solution freezes at -2-55- C- What is its boiling point- K b H 2 O -0-52 K - m - K f H 2 O -1-86 K - m A- 107-0- CB- 100-6- CC- 100-1- CD- 100-7- C

The correct option is D 100.7^∘ C Δ T_f = K_f.m 2.55 = 1.86 × m m = 1.37 Δ T_b = K_b.m = 0.52 × 1.37 = 0.712 T_b = T_b^∘ + Δ T_b = 100 + 0.712 = 100.7^∘ C ...

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Byju's Answer

Standard XII

Chemistry

Relative Lowering of Vapour Pressure

An aqueous so...

Question

An aqueous solution freezes at $- {2.55}^{\circ} C$ . What is its boiling point?

$(K_{b} H_{2} O = 0.52 K / m; K_{f} H_{2} O = 1.86 K / m)$

${107.0}^{\circ} C$

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${100.6}^{\circ} C$

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${100.1}^{\circ} C$

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${100.7}^{\circ} C$

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Solution

The correct option is D
${100.7}^{\circ} C$

$Δ T_{f} = K_{f} . m$

$2.55 = 1.86 \times m$

$m = 1.37$

$Δ T_{b} = K_{b} . m$

$= 0.52 \times 1.37$

$= 0.712$

$T_{b} = T_{b}^{\circ} + Δ T_{b}$

$= 100 + 0.712 = {100.7}^{\circ} C$

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Similar questions

Q. An aqueous solution freezes at

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(K_{b} (H_{2} O) = 0.52 K / m; K_{f} (H_{2} O) = 1.86 k / m)

Q. An aqueous solution freezes at

- {2.55}^{o} C

. What is its boiling point?

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An aqueous solution freezes at −2.55∘ C. What is its boiling point? (Kb H2O=0.52K/m; Kf H2O=1.86K/m)

The correct option is D 100.7∘ C ΔTf=Kf.m 2.55=1.86×m m=1.37 ΔTb=Kb.m =0.52×1.37 =0.712 Tb=T∘b+ΔTb =100+0.712=100.7∘ C

An aqueous solution freezes at $- {2.55}^{\circ} C$ . What is its boiling point?

$(K_{b} H_{2} O = 0.52 K / m; K_{f} H_{2} O = 1.86 K / m)$

The correct option is D
${100.7}^{\circ} C$

$Δ T_{f} = K_{f} . m$

$2.55 = 1.86 \times m$

$m = 1.37$

$Δ T_{b} = K_{b} . m$

$= 0.52 \times 1.37$

$= 0.712$

$T_{b} = T_{b}^{\circ} + Δ T_{b}$

$= 100 + 0.712 = {100.7}^{\circ} C$