How much amount of NaCl should be added to 600 g of water p -1-00 g - mL to decrease freezing point of water to -0-2- C - The freezing point depression constant for water -2 K Kg mol -1

NaCl is strong electrolyte and gives 2 ions in the solution, this implies, i = 2. Molality = w × 1000/58.5 × 600 T_f = 0.2 ^∘C T_f = i × k_f × m On solving we g ...

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

Byju's Answer

Standard XII

Chemistry

Liquids in Liquids

How much amou...

Question

How much amount of NaCl should be added to 600 g of water $(ρ = 1.00 g / m L)$ to decrease freezing point of water to $- 0.2^{\circ} C$ ?
(The freezing point depression constant for water = 2 K Kg $m o l^{- 1}$

Open in App

Solution

NaCl is strong electrolyte and gives 2 ions in the solution, this implies, i = 2.
Molality = $\frac{w \times 1000}{58.5 \times 600}$
$△ T_{f} = 0.2^{\circ} C$
$△ T_{f} = i \times k_{f} \times m$
On solving we get,
w = 1.76 grams

Suggest Corrections

Similar questions

Q. The molal freezing point constant for water is 1.86 K/m. Therefore, the depression in freezing point of 0.1 M

N a C l

solution in water is expected to be:

Q. What weight of glycerol should be added to

600 g

of water in order to lower its freezing point by

10^{o} C

?
Given that:

K_{f} = {1.86}^{o} C . m^{- 1}

Q. Calculate the amount of

K C l

which must be added to 1 kg of water so that the freezing point is depressed by 2K.

(k_{f}

for water

= 1.86 K k g m o l^{- 1})

Q. The freezing point of a diluted milk sample is found to be

- {0.2}^{o} C

, while it should have been

- {0.5}^{o} C

for pure milk. How much water has been added to pure milk. How much water has been added to pure milk to make the diluted sample?

The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water is (molal depression constant = 1.86)

Join BYJU'S Learning Program

How much amount of NaCl should be added to 600 g of water (ρ=1.00g/mL) to decrease freezing point of water to −0.2 ∘C? (The freezing point depression constant for water = 2 K Kg mol−1

NaCl is strong electrolyte and gives 2 ions in the solution, this implies, i = 2. Molality = w×100058.5×600 △Tf=0.2 ∘C △Tf=i×kf×m On solving we get, w = 1.76 grams

How much amount of NaCl should be added to 600 g of water $(ρ = 1.00 g / m L)$ to decrease freezing point of water to $- 0.2^{\circ} C$ ?
(The freezing point depression constant for water = 2 K Kg $m o l^{- 1}$

NaCl is strong electrolyte and gives 2 ions in the solution, this implies, i = 2.
Molality = $\frac{w \times 1000}{58.5 \times 600}$
$△ T_{f} = 0.2^{\circ} C$
$△ T_{f} = i \times k_{f} \times m$
On solving we get,
w = 1.76 grams