Chemistry, asked by varavaralaxmi16, 28 days ago

Calculate the Molality of the solution containing 8gms of NAOH present in 1kg of solution

Answers

Answered by tejindersinghdeol41

Answer:

8g NaOH, mol of NaOH =

8g NaOH, mol of NaOH = 40

8g NaOH, mol of NaOH = 408

8g NaOH, mol of NaOH = 408 =0.2mol

8g NaOH, mol of NaOH = 408 =0.2mol18g of H

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O=

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 18

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH =

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH = 1.2

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH = 1.20.2

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH = 1.20.2 =0.167

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH = 1.20.2 =0.167Molality =

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH = 1.20.2 =0.167Molality = 18

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH = 1.20.2 =0.167Molality = 180.2×1000

8g NaOH, mol of NaOH = 408 =0.2mol18g of H 2 O, mol of H 2 O= 1818 =1 mol∴X NaOH = 1.20.2 =0.167Molality = 180.2×1000 =11.11m.

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