The boiling point of water at 735 torr is 99 -C- The mass of NaCl added in 100 gm of water Kb - 0-51 K kg mol-3 to make its boiling point at 100 -C is-

The correct option is A 5.34 gGive: #160; K_b=0.51 K kg mol^ 1 Δ T_f=100 99=1 K Using the relation, Δ T_f=K_f× m ⇒ m=10.51=1.96 ⇒ n= ...

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Standard XII

Chemistry

Elevation in Boiling Point

The boiling p...

Question

The boiling point of water at 735 torr is 99 $^{\circ} C$ . The mass of $N a C l$ added in 100 gm of water ( $K_{b} = 0.51 K k g m o l^{- 3}$ ) to make its boiling point at 100 $^{\circ} C$ is:

10.68 g

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5.34 g

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2.67 g

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26.7 g

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Solution

The correct option is A 5.34 g
Give: $K_{b} = 0.51 K k g m o l^{- 1}$
$Δ T_{f} = 100 - 99 = 1 K$

Using the relation, $Δ T_{f} = K_{f} \times m$
$\Rightarrow m = \frac{1}{0.51} = 1.96$

$\Rightarrow n = 0.5 \times 1.96 \times 100 \times 10^{- 3} = 0.098$

$\Rightarrow m a s s = 0.098 \times 58.5 = 5.34 g$

Hence, option $B$ is the correct answer.

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The boiling point of water at 735 torr is 99.07°C.The mass of NaCl added in 100 g water to make its boiling point 100°C is ( K_b = 0.51 K kg mol^-1)

ans: 5.34 g

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Q. At higher altitude, boiling point of water is

95^{o} C

. The mass in gram of

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(

α = 0.8

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(K_{b} = 0.52 {K kg mol}^{- 1})

in order to raise the boiling point to

100^{o} C

is:

(M.W of N a C l = 58.5 u)

117 g

N a C l

is added to

222 g

of water in a container. Calculate the temperature at which the water boils at 1 atm?
Given:

Ebullioscopic constant for water (K_{b}) = 0.52 K k g {mol}^{- 1}

Molecular mass of NaCl = 58.5 g m o l^{- 1}

and boiling point of pure water is

100^{\circ} C

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C. Calculate the boiling point of an aqueous solution containing

0.6 g

of urea (molar mass =

60 {g mol}^{- 1}

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K_{b}

for water is

0.52 {K kg mol}^{- 1}

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The boiling point of water at 735 torr is 99 ∘C. The mass of NaCl added in 100 gm of water (Kb=0.51K kg mol−3) to make its boiling point at 100∘C is:

The correct option is A 5.34 gGive: Kb=0.51 K kg mol−1ΔTf=100−99=1 KUsing the relation, ΔTf=Kf×m⇒m=10.51=1.96⇒n=0.5×1.96×100×10−3=0.098⇒ mass=0.098×58.5=5.34 gHence, option B is the correct answer.

The boiling point of water at 735 torr is 99 $^{\circ} C$ . The mass of $N a C l$ added in 100 gm of water ( $K_{b} = 0.51 K k g m o l^{- 3}$ ) to make its boiling point at 100 $^{\circ} C$ is: