What is the molarity of a solution with % W / W = 16 39 0 0 % and density of solution = 1.6 g / ml if molar mass of solute is 78 g / mol ?

Mass of solute = 39 g Volume of solution = 1.6 160 = 100 ml ∴ Molarity = 100 × 1 0 − 3 ( 78 39 ) = 5 M

What is the molarity of a solution with % W / W = 16 39 0 0 % and density of | TestHub

ChemistryMediumnumerical

Passage / Comprehension

The concentrations of solutions can be expressed in number of ways; viz : mass fraction of solute (or mass percent), Molar concentration (Molarity) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each other i.e. knowing one concentration term for the solution, we can find other concentration terms also. The definition of different concentration terms are given below :

Molarity : It is number of moles of solute present in one litre of the solution.

Molality : It is the number of moles of solute present in one kg of the solvent

Mole Fraction = $= \frac{m o l es \circ f so l u t e}{m o l es \circ f so l u t e + m o l es \circ f so l v e n t}$

If molality of the solution is given as 'a' then mole fraction of the solute can be calculated by Mole Fraction = $= \frac{a}{a + \frac{1000}{M _{so l v e n t}}} = \frac{a \times M _{so l v e n t}}{( a \times M _{so l v e n t} + 1000 )}$

where a = molality and M_solvent = Molar mass of solvent

What is the molarity of a solution with $% W / W = \frac{39 0 _{0}}{16} %$ and density of solution $= 1.6 g / ml$ if molar mass of solute is $78 g / mol$ ?

Answer:

5.00

Solution:

Mass of solute $= 39 g$

$Volume of solution = \frac{160}{1.6} = 100 ml ∴ Molarity = \frac{( \frac{39}{78} )}{100 \times 1 0 ^{- 3}} = 5 M$

Stream:JEESubject:ChemistryTopic:Concentration TermsSubtopic:Problems Based on Mixing of Solutions and Dilution

⏱ 2mℹ️ Source: QB

Chemistry - Concentration Terms Question with Solution | TestHub

Doubts & Discussion