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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Calculate the molarity of the following solution: 30 g of Co(NOA3)A2⋅6HA2O in 4.3 L of solution. - Chemistry

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प्रश्न

Calculate the molarity of the following solution: 

30 g of \[\ce{Co(NO3)2 * 6H2O}\] in 4.3 L of solution.

संख्यात्मक
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उत्तर

Molar mass of \[\ce{Co(NO3)2 * 6H2O}\]

= 58.7 + 2(14 + 48) + 6 × 18 g mol−1

= 58.7 + 2(62) + 108 g mol−1

= 58.7 + 124 + 108 g mol−1

= 290.7 g mol−1

∴ Number of moles of \[\ce{Co(NO3)2 * 6H2O}\] = `"Mass"/"Molar mass"`

= `(30  "g")/(290.7  "g mol"^(-1))`

= 0.103 mol

Volume of solution = 4.3 L

Molarity of solution = `"Number of moles of solute"/"Volume of solution in L"`

= `(0.103  "mol")/(4.3 "L")`

= 0.024 M

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अध्याय 1: Solutions - Intext Questions [पृष्ठ ५]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
अध्याय 1 Solutions
Intext Questions | Q 1.3 (a) | पृष्ठ ५

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