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0.58 g of CH3(CH2)(c)COOH was burnt in e...

0.58 g of `CH_3(CH_2)_(c)COOH` was burnt in excess air and the resulting gases `(CO_2` and `H_2O`) were passed through excess NaOH solution. The resulting solution was divided into two equal parts. One part requires 50 " mL of " 1.0 M HCl for complete neutralisation using phenolphthalein indicator. Another part required 80 " mL of " same HCl for neutralisation using methyl orange as indicator. Calculate the value of n and the amount of excess NaOH solution taken initially.

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1.7225 g of a metal (bivalent) salt A_(x)(CO_3)_(y)(OH)_(z) was dissolved in water to make 100 " mL of " solution 50 " mL of " this solution required 10 " mL of " 0.5 M H_2SO_4 solution for complete neutralisation using phenolphthalein indicator. Another 50 mL solution required 15 " mL of " same acid using methyl orange indicator. Deduce the formula of the salt.

1.7225 g of a metal (bivalent) salt A_(x)(CO_3)_(y)(OH)_(z) was dissolved in water to make 100 " mL of " solution 50 " mL of " this solution required 10 " mL of " 0.5 M H_2SO_4 solution for complete neutralisation using phenolphthalein indicator. Another 50 mL solution required 15 " mL of " same acid using methyl orange indicator. Deduce the formula of the salt.

0.1g of a solution containing Na_(2)CO_(3) and NaHCO_(3) requires 10mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator, mass% of Na_(2)CO_(3) in solution is:

0.1g of a solution containing Na_(2)CO_(3) and NaHCO_(3) requires 10mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator, mass% of Na_(2)CO_(3) in solution is: