20 cm\(^3\) of a 2.0 M solution of ethanoic acid was added to excess of 0.05 M sodium hydroxide. The mass of the salt produced is?
(Na = 23, C = 12, O = 16, H = 1)
2.50 g
2.73 g
3.28 g
4.54 g
Explanation
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Discussions (16)
CH3COOH + NaOH -- CH3COONa + H2O ( equation of reaction). ratio of acid to base is 1 : 1 CAVA/CBVB = na/nb; (2.0 × 20)/0.05 × VB = 1/1 ; VB = (2.0 × 20 × 1)/(0.05 × 1) ; VB = 800cm^3. and Molar mass of CH3COONa ( which is d salt produced) = 12 + 3 + 12 + 16 + 16 + 23 = 82. therefore Solubility = mass/ molar mass × 1000/ volume ; 0.05 = mass/ 82 × 1000/ 800 ; mass = (0.05 × 82 × 800)/ 1000 ; mass = 3.28 g option C.
it is simple in diz way....luk at d equation has DBRANDY rightly stated it. u can see dat 1mole of ethanoic acid will produce 1mole of cH3cooNa...meanwhile ethanoic acid has 2mole perdm³ i.e2mole in 1000cm³....in 20cm³=20×2÷1000=0.04mole....acording to d ratio aforemention....dat means dat 0.04mole of ethanoic acid will produce 0.04mole of cH3cooNa....mass =mole ×molarmass....0.04×82(which is d molar mass of d salt)=3.28
