Question

A solution of sodium sulphate 92g of \[N{a^ + }\] ions per kilogram of water. The molality of \[N{a^ + }\] ions in the solution in mole \[k{g^{ - 1}}\]is:
A. 16
B. 8
C. 4
D. 12

Answer 1

Hint:This question is based on two formulas that is Molality and number of moles
1.Number of moles = weight / molecular weight
2.Molality = number of moles of solute / mass of solvent in Kg
First find out the number of moles of \[N{a^ + }\] ion from the number of mole formula and then find out molality.

Complete step by step answer:
Given,
Weight of solute (\[N{a^ + }\] ion) = 92gm
Weight of solvent (water) = 1Kg
We know, molecular weight of \[Na\] is 23g/mole
Find out number of moles of \[N{a^ + }\]:
Number of moles of sodium ion \[N{a^ + }\] = weight of sodium ion \[N{a^ + }\]/ molecular weight of sodium
= 92 gm/ 23\[{\text{gmmo}}{{\text{l}}^{{\text{ - 1}}}}\]
= 4 moles.
Find out molality of sodium ions \[N{a^ + }\]:
Molality is the number of moles of solute present in per Kg of solvent.
Molality of sodium ions = number of moles of \[N{a^ + }\]ions / weight of solvent in Kg
= 4 moles / 1Kg
= 4 moles/kg

So, molality of \[N{a^ + }\] ions in solution is 4\[{\text{molK}}{{\text{g}}^{{\text{ - 1}}}}\].
Hence, the correct option is C.

Note:
Solution is formed by mixing solute and solvent.
In solution how much solute or solvent is present can be calculated by different concentration terms
For ex. Molality, Molarity, Normality, Mole fraction, Mass percent, Volume percent, etc.
Molarity = number of moles of solute / volume of solution in L
Molality = number of moles of solute / mass of solvent in Kg
Mole fraction = number of moles of component whose mole fraction is calculating / number of moles of all components.
Normality = number of gm equivalent of solute / volume of solution in L.
Out of which some concentration terms depend upon temperature and some are independent on temperature
Molality and mole fraction is independent on temperature while molarity and normality depends on temperature.