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Chapter 11: Q41 E (page 648)

A 13.0% solution of \({\bf{K_2CO_3}}\) by mass has a density of 1.09\({\bf{g/g}}{{\bf{m}}^{\bf{3}}}\). Calculate the molality of the solution.

Short Answer

Expert verified

The molality of \({\bf{K_2CO_3}}\) = 1.08 molal

Step by step solution

01

Molality

Molality may be defined as the number of moles of the solute present in the solvent in kilogram. It depends upon the concentration of the solution.

02

Molality of Glucose

13.0% solution means 13 g of \({\bf{K_2CO_3}}\) is in the 100 g solution

Solutionmass = Solute mass+ solvent mass

100 g = 13 g + SolventWeight

Solvent weight = 100 g - 13 g = 87 g

Number of Moles of \({\bf{K_2CO_3 = }}\frac{{{\bf{13}}}}{{{\bf{138}}{\bf{.2}}}}{\bf{ = 0}}{\bf{.094}}\)

Convert the massof solvent into grams

87g = 0.087kg

\({\bf{Molality = }}\frac{{{\bf{Numberofmoles}}}}{{{\bf{MassofSolvent(kg)}}}}\)

Molality of\(K_2CO_3 = \frac{{0.094}}{{0.087}} = 1.08molal\)

Thus, the molality of the answeris 1.08 m

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