Question

The molality of a urea solution in which $ 0.0100g $ of urea. $ [{(N{H_2})_2}CO] $ is added to $ 0.3000d{m^3} $ of water at $ STP $ is:
 $ A)3.33 \times {10^{ - 2}}m \\
  B)0.555m \\
  C)5.55 \times {10^{ - 4}}m \\
  D)3.33m $

Answer 1

Hint : Molality is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. The $ SI $ unit for molality is $ \dfrac{{mol}}{{kg}} $ . In the case of solutions with more than one solvent, molality can be defined for the mixed solvent considered as a pure pseudo solvent.

Complete Step By Step Answer:
Molality is a measure of the number of moles of solute in a solution corresponding to $ 1kg $ or $ 1000g $ of solvent. This contrasts with the definition of molarity which is based on a specified volume of solution. A commonly used unit for molality is $ \dfrac{{mol}}{{kg}} $ . A solution of concentration $ 1\dfrac{{mol}}{{kg}} $ is also sometimes denoted as $ 1molal $ .
 $ \Rightarrow $ $ m = \dfrac{{mol}}{{kg}} $
Where,
 $ M = $ Molality
 $ n = $ Moles of solute
 $ v = $ Kilogram of solvent
We have weight of urea $ = 0.0100g $
Molecular mass of urea $ = 60gmo{l^{ - 1}} $
Therefore, number of moles $ = \dfrac{{0.0100}}{{60.06}} $
 $ = 1.67 \times {10^{ - 4}}mol $
Now, we know mass $ = $ volume $ \times $ Density
One formula we need to know is of the formula for density, which is $ d = \dfrac{m}{v} $ , where d is density, $ m $ is mass and $ v $ is volume
Density of water $ = {10^3}\dfrac{g}{{d{m^3}}} $ and volume $ = 0.3000d{m^3} $
Mass of water $ = 0.3000 \times {10^3} = 300g = 0.3kg $
 $ \Rightarrow $ $ molality = \dfrac{{moles\;of\;solute}}{{mass\;of\;solvent}} $
 $ = \dfrac{{1.67 \times {{10}^{ - 4}}}}{{0.3}} $
 $ = 5.57 \times {10^{ - 4}}m $
So, the correct answer is $ C)5.57 \times {10^{ - 4}} $ .

Note :
Molality is utilized any time you expect the solute may communicate with the solute and in the accompanying circumstances: To decide a limit. To decide a softening point. When working with colligative properties. Molality is the favored concentration transmission approach because the solution's mass of solute and solvent does not change. Another advantage of molality is the fact that the molality of one solute in a solution is independent of the presence or absence of other solutes.