At higher altitude- boiling point of water is 95 oC- The mass in gram of NaCl - -0-8 added to 100 g of water Kb -0-52 K kg mol-1 in order to raise the boiling point to 100 oC is- M-W of NaCl -58-5 u

We know, nbsp; Δ T_B =iK_bm Where, m =molality Given, α =0.8 So, i =1+α =1.8 ∴ 5=1.8× nbsp;0.52×1000100×W58.5 ⇒ W = 5× 100× 58.51.8× 0.52× 1000= 31.25 g ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Definition of Solution

At higher alt...

Question

At higher altitude, boiling point of water is $95^{o} C$ . The mass in gram of $N a C l$ ( $α = 0.8$ ) added to $100 g$ of water $(K_{b} = 0.52 {K kg mol}^{- 1})$ in order to raise the boiling point to $100^{o} C$ is:
$(M.W of N a C l = 58.5 u)$

Open in App

Solution

We know,
$Δ T_{B} = i K_{b} m$
Where, $m = molality$
Given, $α = 0.8$
So, $i = 1 + α = 1.8$
$∴ 5 = 1.8 \times 0.52 \times \frac{1000}{100} \times \frac{W}{58.5}$
$\Rightarrow W = \frac{5 \times 100 \times 58.5}{1.8 \times 0.52 \times 1000} = 31.25 g$

Suggest Corrections

Similar questions

Q. 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:

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

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

how the answer came

I want step by step detailed explanation

Q. The boiling point of pure water is

100^{\circ}

C. Calculate the boiling point of an aqueous solution containing

0.6 g

of urea (molar mass =

60 {g mol}^{- 1}

) in

100 g

of water.

K_{b}

for water is

0.52 {K kg mol}^{- 1}

10 g

of a non-electrolyte solute

(M = 60 g/mol)

is added to

250 g

of water. The boiling point of the solution is (

K_{b}

for water is

0.52 {K kg mol}^{- 1}

and boiling point of water

= 373.15 K

)
[0.77 Mark]

Join BYJU'S Learning Program

At higher altitude, boiling point of water is 95 oC. The mass in gram of NaCl (α=0.8) added to 100 g of water (Kb=0.52 K kg mol−1) in order to raise the boiling point to 100 oC is: (M.W of NaCl=58.5 u)

We know, ΔTB=iKbm Where, m=molality Given, α=0.8 So, i=1+α=1.8 ∴5=1.8×0.52×1000100×W58.5 ⇒W=5×100×58.51.8×0.52×1000=31.25 g

At higher altitude, boiling point of water is $95^{o} C$ . The mass in gram of $N a C l$ ( $α = 0.8$ ) added to $100 g$ of water $(K_{b} = 0.52 {K kg mol}^{- 1})$ in order to raise the boiling point to $100^{o} C$ is:
$(M.W of N a C l = 58.5 u)$

We know,
$Δ T_{B} = i K_{b} m$
Where, $m = molality$
Given, $α = 0.8$
So, $i = 1 + α = 1.8$
$∴ 5 = 1.8 \times 0.52 \times \frac{1000}{100} \times \frac{W}{58.5}$
$\Rightarrow W = \frac{5 \times 100 \times 58.5}{1.8 \times 0.52 \times 1000} = 31.25 g$