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Simran Bhatia , 11 Years ago
Grade 11
anser 4 Answers
Radhika Batra

Last Activity: 11 Years ago

0.375 M aqueous solution of sodium acetate

≡ 1000 mL of solution containing 0.375 moles of sodium acetate

∴Number of moles of sodium acetate in 500 mL

 

Molar mass of sodium acetate = 82.0245 g mole–1 (Given)

∴ Required mass of sodium acetate = (82.0245 g mol–1) (0.1875 mole)

= 15.38 g

 

Lovish singla

Last Activity: 7 Years ago

Molarity°mass/volume; put molarity to be0.375 and volume to be 500 ml we will get moles to be .188 and then multiply by 82 mass of sodium acetate

aditya pradeep

Last Activity: 6 Years ago

0.375M aqueous solution means that 0.375 mo of sodium acetate are present in 1000mL of solution.500mL of the solution should contain sodium acetate.
                                                                         =0.375/2 mole
 molar mass of sodium acetate   =82.0245 gmole-1 
mass of sodium acetate required=(0.375/2) (82.0245)
                                                    =15.38 g 
 

ankit singh

Last Activity: 4 Years ago

0.375 M solution of CH3​COONa = 0.375 moles of CH3​COONa/1000ml

No. of moles of CH3​COONa in 500ml =10000.375​×500=0.1875 moles
Molar mass of sodium acetate = 82.0245 g/mole
∴ mass of CH3​COONa required =82.0245×0.1875=15.38g

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