Chemistry - Concentration Terms Question with Solution | TestHub

ChemistryConcentration TermsConcentration termMedium2 minPYQ_2024

ChemistryMediumnumerical

The mass of sodium acetate $({CH}_{3} COONa)$ required to prepare $250 mL$ of $0.35 M$ aqueous solution is _____ g. $($ Molar mass of ${CH}_{3} COONa$ is $82.02 g {mol}^{- 1}$ ) Round off to the nearest integer.

Answer:

7.00

Solution:

Given, Molarity= $0.35 M$

Moles $=$ Molarity $\times$ Volume in litres

$= 0.35 \times 0.25$

No. of moles = Mass/ Molar mass

Mass $=$ moles $\times$ molar mass

$= 0.35 \times 0.25 \times 82.02 = 7.18 g$

Ans. $7$

Stream:JEESubject:ChemistryTopic:Concentration TermsSubtopic:Concentration term

⏱ 2mℹ️ Source: PYQ_2024

Doubts & Discussion

Loading discussions...