When some $N a C l$ was dissolved in water, the freezing point depression was numerically equal to twice the molal $K_{f}$ depression constant. The relative lowering of vapour pressure of the solution is nearly:

0.036

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.018

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0585

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.072

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0.036$
$Δ T_{f} = K_{f} \frac{W_{1}}{M_{1} \times W_{2}} \times 1000$ [ $W_{2}$ is given in grams]
where $1$ and $2$ denote solute & solvent respectively.
Given, $Δ T_{f} = 2 K_{f} \Rightarrow \frac{W_{1} \times 1000}{M_{1} \times W_{2}} = 2$
Relative lowering of vapour pressure= $\frac{W_{1} \times M_{2}}{M_{1} \times W_{2}}$
Here, $M_{2}$ = Molecular weight of $H_{2} O = 18$
$∴$ Relative lowering of vapour pressure= $(\frac{W_{1}}{M_{1} \times W_{2}}) \times 18$
$= \frac{2}{1000} \times 18$
$= 0.036$

Suggest Corrections

Similar questions

Question 53
Assertion (A) When NaCl is added to water a depression in freezing point is observed.
Reason (R) The Lowering of vapour pressure of a solution causes depression in the freezing point.

1.00 g

of a non-electrolyte solute (molar mass

250 g m o l^{- 1}

) was dissolved in

51.2 g

of benzene. If the freezing point depression constant,

K_{f}

of benzene is

5.12 K k g m o l^{- 1}

, the freezing point of benzene will be lowered by:

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

Q. 0.534 g of solute is dissolved in 15 g of water then freezing point temperature changes from

0^{o} C

- {1.57}^{o} C

. Molal depression constant of water,

K_{f} = 1.85 K k g m o l^{- 1}

. Find out:
(i) Molal concentration
(ii) Molecular mass of solute

Join BYJU'S Learning Program

When some NaCl was dissolved in water, the freezing point depression was numerically equal to twice the molal Kf depression constant. The relative lowering of vapour pressure of the solution is nearly:

When some $N a C l$ was dissolved in water, the freezing point depression was numerically equal to twice the molal $K_{f}$ depression constant. The relative lowering of vapour pressure of the solution is nearly: